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Question: If \(\cos 5\theta = a\cos \theta + b{\cos ^3}\theta + c{\cos ^5}\theta + d\), then \(A)a = 20\) ...

If cos⁡5θ=acos⁡θ+bcos⁡3θ+ccos⁡5θ+d\cos 5\theta = a\cos \theta + b{\cos ^3}\theta + c{\cos ^5}\theta + d, then
A)a=20A)a = 20
B)b=−20B)b = - 20
C)c=16C)c = 16
D)d=5D)d = 5

Explanation

Solution

This problem comes under trigonometry, this question has to find the values of variables with which we need to compare the values with answers given in the option and we need to find the correct one. Here we use some trigonometric identities and basic algebraic identities and basic mathematical calculation and complete step by step explanation.

Formula used: cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B
cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2{\cos ^2}\theta - 1
cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta
sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta
(a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta

Complete step-by-step solution:
Now consider the question cos⁡5θ\cos 5\theta
Now separate the 5θ5\theta into 2θ+3θ2\theta + 3\theta
cos⁡(2θ+3θ)\cos (2\theta + 3\theta )
By using formula mentioned in formula used, we get
cos⁡(2θ+3θ)=cos⁡2θcos⁡3θ−sin⁡2θsin⁡3θ\cos (2\theta + 3\theta ) = \cos 2\theta \cos 3\theta - \sin 2\theta \sin 3\theta
Again, using formulas for the above values separately mentioned in formula used and by substituting the values, we get
⇒(2cos⁡2θ−1)(4cos⁡3θ−3cos⁡θ)−2sin⁡θcos⁡θ(3sin⁡θ−4cos⁡3θ)\Rightarrow (2{\cos ^2}\theta - 1)(4{\cos ^3}\theta - 3\cos \theta ) - 2\sin \theta \cos \theta (3\sin \theta - 4{\cos ^3}\theta )
By using algebraic multiplication, we get
⇒8cos⁡5θ−6cos⁡3θ−4cos⁡3θ+3cos⁡θ−6sin⁡2θcos⁡θ+8sin⁡4θcos⁡θ\Rightarrow 8{\cos ^5}\theta - 6{\cos ^3}\theta - 4{\cos ^3}\theta + 3\cos \theta - 6{\sin ^2}\theta \cos \theta + 8{\sin ^4}\theta \cos \theta
Now converting, we get
⇒8cos⁡5θ−6cos⁡3θ−4cos⁡3θ+3cos⁡θ−6sin⁡2θcos⁡θ+8(sin⁡2θ)2cos⁡θ\Rightarrow 8{\cos ^5}\theta - 6{\cos ^3}\theta - 4{\cos ^3}\theta + 3\cos \theta - 6{\sin ^2}\theta \cos \theta + 8{({\sin ^2}\theta )^2}\cos \theta
Once again using formula mentioned in formula used, we get
⇒8cos⁡5θ−6cos⁡3θ−4cos⁡3θ+3cos⁡θ−6(1−cos⁡2θ)cos⁡θ+8(1−cos⁡2θ)2cos⁡θ\Rightarrow 8{\cos ^5}\theta - 6{\cos ^3}\theta - 4{\cos ^3}\theta + 3\cos \theta - 6(1 - {\cos ^2}\theta )\cos \theta + 8{(1 - {\cos ^2}\theta )^2}\cos \theta
Similarly,
⇒8cos⁡5θ−6cos⁡3θ−4cos⁡3θ+3cos⁡θ−6(cos⁡θ−cos⁡3θ)+8(1+cos⁡4θ−2cos⁡2θ)cos⁡θ\Rightarrow 8{\cos ^5}\theta - 6{\cos ^3}\theta - 4{\cos ^3}\theta + 3\cos \theta - 6(\cos \theta - {\cos ^3}\theta ) + 8(1 + {\cos ^4}\theta - 2{\cos ^2}\theta )\cos \theta
By multiplying the numerals in the to obtain the equation, we get
⇒8cos⁡5θ−6cos⁡3θ−4cos⁡3θ+3cos⁡θ−6cos⁡θ−6cos⁡3θ+8(cos⁡θ+cos⁡5θ−2cos⁡3θ)\Rightarrow 8{\cos ^5}\theta - 6{\cos ^3}\theta - 4{\cos ^3}\theta + 3\cos \theta - 6\cos \theta - 6{\cos ^3}\theta + 8(\cos \theta + {\cos ^5}\theta - 2{\cos ^3}\theta )
By adding the same term coefficient, we get
⇒8cos⁡5θ−6cos⁡−4cos⁡3θ+3cos⁡θ−6cos⁡θ+8cos⁡θ+8cos⁡5θ−16cos⁡3θ\Rightarrow 8{\cos ^5}\theta - 6\cos - 4{\cos ^3}\theta + 3\cos \theta - 6\cos \theta + 8\cos \theta + 8{\cos ^5}\theta - 16{\cos ^3}\theta
By eliminating common terms, we get
⇒16cos⁡5θ−20cos⁡3θ+5cos⁡θ\Rightarrow 16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta
By rearranging in the form of given equation, we get
B)b=−20B)b = - 20
By comparing the above equation coefficient with the coefficient of equation given in question, we obtain the values of a, b, c, d.
Therefore the values are a=5,b=−20,c=16,d=0a = 5,b = - 20,c = 16,d = 0

The answer is B)b=−20B)b = - 20

Note: We have to remember that trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. The Greeks focused on the calculation of chords. While mathematicians in India created earliest-known tables of values for trigonometric ratios (also called trigonometric functions) such as sine. Throughout history, trigonometry has been applied in areas such as geodesy, surveying celestial mechanics, and navigation. Trigonometry is known for its many identities, which are equations used for rewriting trigonometric expressions to solve equations, to find a more useful expression, or to discover new relationships.