..READ THIS
FIRST..Problems
about trigonometry
READ THIS
FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.
Problems about trigonometry
Level 1 problems
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Calculate cos(3u) in terms of cos(u) cos(2u+u) = cos(2u)cos(u)-sin(2u)sin(u) = (cos2(u)-sin2(u))cos(u)-2sin(u)cos(u)sin(u) = cos3(u) - sin2(u)cos(u)-2sin2(u)cos(u) = cos3(u) -3sin2(u)cos(u) cos3(u) -3cos(u)+3cos3(u) 4cos3(u) -3cos(u) In the same way sin(3u) = -4sin3(u) -3sin(u)
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prove that : cos(t)+sin(t) cos(2t) ------------- = ------------- cos(t)-sin(t) 1- sin(2t)
cos(2t) ---------- = 1- sin(2t) cos2(t)-sin2(t) (cos(t)-sin(t))(cos(t)+sin(t)) --------------- = ------------------------------------------ = 1-2sin(t)cos(t) cos(t)cos(t)-2sin(t)cos(t)+sin(t)sin(t) (cos(t)-sin(t))(cos(t)+sin(t)) ------------------------------- = (cos(t)-sin(t))(cos(t)-sin(t)) cos(t)+sin(t) ------------ cos(t)-sin(t)
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show that sin(p)-sin(q) p+q -------------- = cot(-----) cos(q)-cos(p) 2
sin(p)-sin(q) -------------- = cos(q)-cos(p) 2cos((p+q)/2)sin((p-q)/2) -------------------------- = -2sin((p+q)/2)sin((q-p)/2) cos((p+q)/2) ------------ = sin((p+q)/2) p+q cot(-----) 2 -
Solve (sin(2a)).x2 - 2.x.(sin(a)+cos(a)) +2 =0
Discriminant = 4(sin(a)+cos(a))2 - 8.sin(2a) = 4.( sin2 (a) + 2.sin(a).cos(a) + cos2 (a) - 4 sin(a)cos(a) ) = 4.(sin(a) - cos(a))2 So, the roots are sin(a) + cos(a) + sin(a) - cos(a) 1 x = --------------------------------- =...= ------ sin(2a) cos(a) 1 and x = ... = ------ sin(a)
Level 2 problems
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A plane is approaching your home, and you assume that it is traveling at approximately 550 miles per hour.
If the angle of elevation of the plane is 16 degrees at one time and one minute later the angle is 57 degrees, approximate the altitude.
We assume that the altitude is constant.
The red line is the plane. First the plane is at point Q and one minute later the plane is at point R. Point H is my home.
The distance |QR| is 550/60 = 55/6 miles.
Let D = the distance |PH|.In triangle QPH, tan(16°) = A/D => D = A.cot(16°) In triangle RSH tan(57°) = A/(D - 55/6) A tan(57°) = -------------------- A.cot(16°) - 55/6 (A.cot(16°) - 55/6 ).tan(57°) = A A.(cot(16°).tan(57°) - 1 ) = (55/6) .tan(57°) A = 3.23 miles.The altitude is 3.23 miles. -
Three real numbers a,b,c are successive terms of an arithmetic sequence. Show that sin(a)+sin(b)+sin(c) --------------------- = tan(b) cos(a)+cos(b)+cos(c)
We can represent the tree terms by b-v,b,b+v . Then we have to show thatsin(b-v)+sin(b)+sin(b+v) ------------------------- = tan(b) <=> cos(b-v)+cos(b)+cos(b+v) 2sin(b)cos(v)+sin(b) --------------------- = tan(b) <=> 2cos(b)cos(v)+cos(b) sin(b)(2cos(v)+1) ----------------- = tan(b) <=> cos(b)(2cos(v)+1) sin(b) ------- = tan(b) cos(b)
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Given: a+b+c=pi Prove that cos(2a)+cos(2b)+cos(2c)+1=-4cos(a)cos(b)cos(c)
cos(2a)+cos(2b)+cos(2c)+1= 2cos(a+b)cos(a-b)+2cos2 (c) = but a+b=pi-c =>cos(a+b)=-cos(c) -2cos(c)cos(a-b)+2cos2(c) = -2cos(c).(cos(a-b)-cos(c)) = -2cos(c).(cos(a-b)-cos(a+b)) = -2cos(c).2cos(a)cos(b) = -4cos(a)cos(b)cos(c) -
Find the period of 3.sin2 (3x/4) - cos2 (2x/3)
2.sin2(3x/4) = 1 - cos(3x/2) and this part has period 4.pi/3 So 3.sin2(3x/4) has period 4.pi/3 (1) 2. cos2(x/3) = 1 + cos(2x/3) and this part has period 3.pi So cos2(x/3) has period 3.pi (2) From (1) and (2) we have the period of 3.sin2 (3x/4) - cos2 (2x/3) is 12.pi
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Prove that tan(pi/8) = sqrt(2) - 1
We know that tan(pi/4) = 1.
Let tan(pi/8) = x .
We know that2 tan(u) tan(2u) = --------------- 1- tan(u)tan(u)Make u = pi/82 x 1 = ------- 1- x.x <=> x2 + 2x - 1 = 0 Since tan(pi/8) is positive, we only take the positive root. x = sqrt(2) - 1 -
a+b+c=pi . Prove that tan(a)+tan(b)+tan(c) =tan(a).tan(b).tan(c) b+c = pi - a => tan(b+c) = tan(pi - a) tan(b) + tan(c) => ----------------- = -tan(a) 1 - tan(b).tan(c) => tan(b) +tan(c) = -tan(a).(1 - tan(b).tan(c)) => tan(a)+tan(b)+tan(c) =tan(a).tan(b).tan(c) -
Prove that sin(7.pi/12) = (sqrt(6) +sqrt(2))/4
sin(7.pi/12) = sin (pi/3 + pi/4) = sin(pi/3).cos(pi/4) + cos(pi/3).sin(pi/4) = sqrt(3)/2 .sqrt(2)/2 + 1/2 . sqrt(2)/2 = (sqrt(6) +sqrt(2))/4
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a+b+c=pi. Prove that cos(a).cos(a) + cos(b).cos(b) + cos(c).cos(c) + 2cos(a).cos(b).cos(c) =1
a+b = pi - c => cos(a+b) = cos(pi - c) => cos(a+b) = -cos(c) => cos(a).cos(b) - sin(a).sin(b) = - cos(c) => cos(a).cos(b) + cos(c) = sin(a).sin(b) => (cos(a).cos(b) + cos(c))2 = (sin(a).sin(b))2 => (cos(a).cos(b))2 +2.cos(a).cos(b).cos(c) + cos2(c) = (1 - cos2(a)).(1 - cos2(b)) => ... => cos(a).cos(a) + cos(b).cos(b) + cos(c).cos(c) + 2cos(a).cos(b).cos(c) =1 -
Prove that sin(a) + sin(b) + sin(c) = sin(a + b + c) + 4sin((b+c)/2) sin((c+a)/2) sin((a+b)/2)
a + b a - b sin(a) + sin(b) = 2 sin(-----) cos(-----) 2 2 a + b + 2 c a + b sin(a + b + c) - sin(c) = 2 cos ----------- sin(-----) 2 2 Then sin(a)+ sin(b) + sin(c) - sin(a + b + c) a + b a - b a + b + 2 c = 2 sin(-----) (cos(-----) - cos ----------- ) 2 2 2 a + b a + c b + c = 2 sin(-----).2 sin(-----)sin(-----) 2 2 2 b + c c + a a + b = 4sin(-----) sin(-----) sin(-----) 2 2 2 -
Solve: 2.tan(x) = sin(4x) - 2.sin(2x)tan(x)cos(2x)tan(x)<=> 2.tan(x) = sin(4x) - sin(4x).tan2 (x) <=> 2.tan(x) = sin(4x)(1 - tan2(x)) 2.sin(x) sin(4x).(cos2(x) - sin2(x)) <=> --------- = ---------------------------- cos(x) cos(x).cos(x) <=> 2.sin(x). cos(x) = sin(4x).(cos2 (x) - sin2 (x)) and cos(x) not 0 <=> sin(2x) - 2.sin(2x). cos(2x).cos(2x) and cos(x) not 0 <=> sin(2x) .(1 - 2.cos2 (2x)) = 0 and cos(x) not 0 <=> 2.sin(x). cos(x).(1 - 2.cos2 (2x)) = 0 and cos(x) not 0 <=> sin(x) = 0 or cos(2x) = 1/sqrt(2) or cos(2x) =-1/sqrt(2) <=> x = k.pi or 2x = pi/4 + k(pi/2) <=> x = k.pi or x = pi/8 + k(pi/4) -
Prove that sin(2 arctan(x)) = 2x/(1 + x2)
Let arctan(x) = a => tan(a) = x sin(2 arctan(x)) = sin(2a) = 2sin(a)cos(a) = 2 tan(a).cos(a).cos(a) 2 tan(a) = ------------------- 1 + tan(a).tan(a) = 2x/(1 + xx) -
Calculate the range of the function f(x) = arctan(x2+ 1).
The range of (x2+ 1) is [1, infty[ .
Let the arctan function act on that range.
Then, the set of all images is [pi/4, pi/2 [ . The range of f(x) = [pi/4, pi/2 [ .
Level 3 problems
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tan(a) and tan(b) are the roots of x2 + p.x + q = 0 Prove that sin(a+b).sin(a+b) + p.sin(a+b).cos(a+b) + q.cos(a+b).cos(a+b) = q
tan(a) + tan(b) = -p and tan(a).tan(b) = q sin(a+b).sin(a+b) + p.sin(a+b).cos(a+b) + q.cos(a+b).cos(a+b) sin(a+b).sin(a+b) + p.sin(a+b).cos(a+b) + q.cos(a+b).cos(a+b) = ------------------------------------------------------------- sin(a+b).sin(a+b) + cos(a+b).cos(a+b) divide numerator and denominator by cos(a+b).cos(a+b) tan(a+b).tan(a+b)+ p.tan(a+b) +q = --------------------------------- tan(a+b).tan(a+b) + 1 tan(a) + tan(b) tan(a) + tan(b) tan(a) + tan(b) -----------------. --------------- + p ---------------- +q 1 - tan(a)tan(b) 1 - tan(a)tan(b) 1 - tan(a)tan(b) = ---------------------------------------------------------------- tan(a) + tan(b) tan(a) + tan(b) ---------------- . ---------------- + 1 1 - tan(a)tan(b) 1 - tan(a)tan(b) - p - p - p ------. ------ + p. ------ + q 1 - q 1 - q 1 - q = ---------------------------------- - p - p ------ .------ + 1 1 - q 1 - q = .... a little algebra = q -
Given : cos(2x) = a (cos(x) - sin(x)) (a is a real parameter) Solve this equation and discuss.
cos(2x) = a (cos(x) - sin(x)) <=> cos2 (x) - sin2 (x) = a (cos(x) - sin(x)) <=> (cos(x) + sin(x))(cos(x) - sin(x)) = a (cos(x) - sin(x)) <=> (cos(x) - sin(x))(cos(x) + sin(x) -a) = 0 <=> cos(x) = sin(x) or cos(x) + sin(x) -a = 0 <=> x = pi/4 + k.pi or cos(x) + tan(u).sin(x) = a with u = pi/4 <=> x = pi/4 + k.pi or cos(u).cos(x)+sin(u).sin(x)=a.cos(u) with u = pi/4 <=> x = pi/4 + k.pi or cos(x - u) = a/sqrt(2) (*) The last part has solutions if and only if a in [-sqrt(2) , sqrt(2) ] Case 1 : a is not in [-sqrt(2) , sqrt(2) ] The solutions are x = pi/4 + k.pi Case 2 : a in [-sqrt(2) , sqrt(2) ] Now it is possible to choose t = arccos(a/sqrt(2)) then (*) gives x = pi/4 + k.pi or x = pi/4 + t + 2.k.pi or x = pi/4 - t + 2.k.pi -
Prove that for all x > 1 2 arctan(x) + Bgsin ( 2x/ (1 + x2) ) is constant
Let arctan(x) = y. Then, x = tan(y) with y in ]pi/4 , pi/2[ 2x 2 tan(y) -------- = ----------------- = sin(2y) with 2y in ]pi/2 , pi[ 1 + x2 1 + tan(y).tan(y) <=> 2x -------- = sin(pi - 2y) with (pi - 2y) in ]-pi/2 , 0 [ 1 + x2 <=> 2x arcsin( -------- ) = pi - 2y 1 + x2 Hence, 2x 2 arctan(x) + Bgsin -------- = 2y + pi - 2y = pi 1 + x2 -
Solve F = ( 2 sin(2 x) - 1 ) / (cos(2 x) - 3 cos(x) + 2) > 0
with x in [0,2.pi]
First we investigate N = 2 sin(2 x) - 1 2 sin(2 x) - 1 > 0 <=> sin(2 x) > 1/2 <=> pi/6 + 2.k.pi < 2x < 5.pi/6 + 2.k.pi <=> pi/12 + k.pi < x < 5.pi/12 + k.pi Now we investigate D = cos(2 x) - 3 cos(x) + 2 cos(2 x) - 3 cos(x) + 2 > 0 <=> 2 cos2 (x) - 3 sin(x) + 1 > 0 <=> 2(cos(x) - 1/2)(cos(x) - 1) > 0 since (cos(x) - 1) =< 0 we have <=> cos(x) < 1/2 <=> pi/3 + 2.k.pi < x < 5.pi/3 + 2.k.pi Now we see that 1 4 5 13 17 20 --pi --pi --pi --pi --pi --pi 2pi x 0 12 12 12 12 12 12 --------------------------------------------------------------------------- T - - 0 + + + 0 - 0 + 0 - - - - N 0 - - - 0 + + + + + + + 0 - 0 F | + 0 - | + 0 - 0 + 0 - | + | From this table we can easyly find the solutions of F > 0 in [0, 2.pi]. -
The angles of a triangle ABC are a,b and c. Prove that The triangle is rectangular if and only if sin(4a) + sin(4b) + sin(4c) = 0
Part 1: if the triangle is rectangular then sin(4a)+sin(4b)+sin(4c) = 0.
triangle is rectangular => a or b or c is pi/2 Suppose a = pi/2, then sin(4a) = 0 and b = pi/2 -c => 4 b = 2 pi - 4c => sin(4b) = sin(-4c) => sin(4b) + sin(4c) = 0 So sin(4a) + sin(4b) + sin(4c) = 0 Similar proof for b = pi/2 or c = pi/2Part 2: if sin(4a)+sin(4b)+sin(4c) = 0 then the triangle is rectangular.a + b + c = pi => 4a + 4b + 4c = 4 pi => 4a = 4 pi - (4b + 4c) => sin(4a) = sin( -(4b + 4c)) = - sin(4b + 4c) => sin(4a) = - (sin(4b)cos(4c) + cos(4b)sin(4c)) Thus, sin(4a)+sin(4b)+sin(4c) = sin(4b) - sin(4b)cos(4c) + sin(4c) - cos(4b)sin(4c) <=> 0 = sin(4b) (1 - cos(4c)) + sin(4c) (1 - cos(4b)) <=> 0 = sin(4b) 2 sin2 (2c) + sin(4c).2sin2 (2b) <=> 0 = 2.sin(2b).cos(2b).2 sin2 (2c) + 2.sin(2c).cos(2c).2 sin2 (2b) <=> 0 = 4.sin(2b).sin(2c).sin(2b + 2c) <=> sin(2b) = 0 or sin(2c) = 0 or sin(2b + 2c) = 0 <=> b = pi/2 or c = pi/2 or b + c = pi/2 <=> b = pi/2 or c = pi/2 or a = pi/2 -
Solve x 1 1 1 cot(-) - cot(x) = ------ + -------- + -------- 2 sin(x) sin(2 x) sin(4 x) with x in [0, 2pi]x cot(-) - cot(x) 2 x cos(-) 2 cos(x) = ------ - ------ x sin(x) sin(-) 2 x 2 x 2 x cos(-) cos (-) - sin (-) 2 2 2 = ------ - ------------------ x x x sin(-) 2 sin(-) cos(-) 2 2 2 = ... 1 = ----------------- x x 2 sin(-) cos(-) 2 2 1 = -------- sin(x) With this result the equation is 1 1 -------- + -------- = 0 sin(2 x) sin(4 x) 1 1 <=> -------- = - -------- sin(2 x) sin(4 x) <=> sin(2 x) = - sin(4 x) with sin(4 x) not 0 <=> sin(2 x) = sin(- 4 x) with sin(4 x) not 0 <=> 2x = -4x + 2.k.pi or 2x = pi + 4x + 2.k.pi with sin(4 x) not 0 <=> 6x = 2.k.pi or -2x = pi + 2.k.pi with sin(4 x) not 0 <=> x = k.pi/3 or x = -pi/2 - k.pi with sin(4 x) not 0 The solutions in [0, 2pi] are pi/3 ; 2pi/3 ; 4pi/3 ; 5pi/3 -
Solve the equation ___ V 2 (sin(x) + cos(x)) + 2 cos4 (x - pi/4 ) = 0First sin(x) + cos(x) pi = sin(x) + sin(-- - x) 2 pi pi = 2 sin(--) cos(x - ---) 4 4 ___ pi =V 2 cos(x - ---) 4 With this, the equation becomes pi 4 pi 2 cos(x - ---) + 2 cos (x - --- ) = 0 4 4 <=> pi 3 pi cos(x - ---) = 0 or 1 + cos (x - --- ) = 0 4 4 <=> 3 pi pi x = ---- + k.pi or cos(x - ---) = -1 4 4 <=> 3 pi 3 pi x = ---- + k.pi or x = ---- + 2.k.pi 4 4 <=> 3 pi x = ---- + k.pi 4 -
Given: in a triangle ABC with sides a,b and c A - C A + C b cos(-----) - 3 c cos(-----) = 0 2 2 Prove that a = 2 cProof: Appealing to the sine rule we can write A - C A + C b cos(-----) - 3 c cos(-----) = 0 2 2 <=> A - C A + C sin(B) cos(-----) - 3 sin(C) cos(-----) = 0 2 2 Now, A + B + C = pi A + C pi B => (-----) = -- - - 2 2 2 => A + C B cos(-----) = sin - 2 2 Hence A - C A + C sin(B) cos(-----) - 3 sin(C) cos(-----) = 0 2 2 <=> A - C B sin(B) cos(-----) - 3 sin(C) sin - = 0 2 2 <=> B B A - C B 2 sin - cos - cos(-----) - 3 sin(C) sin - = 0 2 2 2 2 <=> B B A - C sin - .(2 cos - cos(-----) - 3 sin(C))= 0 2 2 2 B sin - = 0 is impossible in a triangle, 2 B A + C and cos - = sin(-----) , hence 2 2 A + C A - C <=> 2(sin(-----)cos(-----) = 3 sin(C) 2 2 <=> sin(A) + sin (C) = 3 sin(C) <=> sin(A) = 2 sin(C) <=> a = 2 c -
Given: in a triangle ABC a cos(B) + b cos(C) = A tan(A) tan(B/2) cos(B) + b tan(B/2) sin(C) Prove that the triangle is isosceles.Proof: a cos(B) + b cos(C) = A tan(A) tan(B/2) cos(B) + b tan(B/2) sin(C) <=> a cos(B)(1 - tan(A) tan(B/2)) = b(tan(B/2) sin(C) - cos(C)) <=> a tan(B/2) sin(C) - cos(C) - = ----------------------------- b cos(B)(1 - tan(A) tan(B/2)) sin(A) tan(B/2) sin(C) - cos(C) <=> ------- = --------------------------- sin(B) cos(B)(1 - tan(A) tan(B/2)) <=> sin(A) cos(B)(1 - tan(A) tan(B/2)) =sin(B)(tan(B/2) sin(C) - cos(C)) multiplying through by cos(B/2) <=>sin(A)cos(B)(cos(B/2)-tan(A)sin(B/2))=sin(B)(sin(B/2)sin(C)-cos(C)cos(B/2)) sin(A)cos(B) <=> ------------(cos(A)cos(B/2)-sin(A)sin(B/2)) = -sin(B)cos(C+B/2) cos(A) <=> tan(A)cos(B)cos(A + B/2) = - sin(B)cos(C+B/2) Now (A + B/2) + (C+B/2) = pi => cos(A + B/2) = - cos(C+B/2) <=> tan(A)cos(B)cos(A + B/2) = sin(B)cos(A + B/2) <=> tan(A)cos(B)cos(A + B/2) - sin(B)cos(A + B/2) = 0 <=> cos(A + B/2) = 0 or tan(A)cos(B) - sin(B) = 0 <=> A + B/2 = pi/2 or tan(A)cos(B) = sin(B) <=> 2A + B = pi or tan(A) = tan(B) <=> 2A + B = A + B + C or A = B <=> A = C <=> the triangle is isosceles.
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