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If $\cos 2x + 2\cos x = 1$ then, ${\sin ^2}x(2 - {\cos ^2}x)$ is
$
  {\text{A}}{\text{. 1}} \\
  {\text{B}}{\text{. 2}} \\
  {\text{C}}{\text{. 4}} \\
  {\text{D}}{\text{. none of these}} \\
 $

Answer Verified Verified
Hint: The trigonometric function is the function that relates the ratio of the length of two sides with the angles of the right-angled triangle widely used in navigation, oceanography, the theory of periodic functions, projectiles. Commonly used trigonometric functions are the sine, the cosine, and the tangent, whereas the cosecant, the secant, the cotangent are their reciprocal, respectively.



The value of these functions can be determined by the relation of the sides of a right-angled triangle where sine function is the ratio of perpendicular (P) and hypotenuses (H) of the triangle\[\sin \theta {\text{ = }}\dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuses}}}}\]. Cosine is the ratio of the base (B) and hypotenuses (H) of the triangle\[\cos \theta {\text{ = }}\dfrac{{{\text{base}}}}{{{\text{hypotenuses}}}}\]; the tangent is the ratio of the perpendicular (P) and base (B) of the triangle \[\tan \theta {\text{ = }}\dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}\] , whereas${\text{cosec }}\theta $ , \[\sec {\text{ }}\theta \] and \[\cot {\text{ }}\theta \] are their inverse respectively. The given value determines the value of these functions.

In this question, trigonometric identities have been used to simplify the question and get the result. We try to get the same trigonometric functions everywhere so as to get the desired result.
Some of the trigonometric identities are:

$
  {\sin ^2}x + {\cos ^2}x = 1 \\
  {\sec ^2}x - {\tan ^2}x = 1 \\
  {\text{cose}}{{\text{c}}^2}x - {\cot ^2}x = 1 \\
  1 - \cos 2x = 2{\sin ^2}x \\
  1 + \cos 2x = 2{\cos ^2}x \\
 $

Complete step by step answer:
The given equation $\cos 2x + 2\cos x = 1$ can be re-written as:
$
  \cos 2x + 2\cos x = 1 \\
  2{\cos ^2}x - 1 + 2\cos x = 1 \\
  2{\cos ^2}x + 2\cos x - 2 = 0 \\
  {\cos ^2}x + \cos x - 1 = 0 \\
  {\cos ^2}x + \cos x = 1 - - - - (a) \\
  1 - {\cos ^2}x = \cos x \\
  {\sin ^2}x = \cos x - - - - (i) \\
 $


Again, the equation ${\sin ^2}x(2 - {\cos ^2}x)$ can be re-written as:

$
  {\sin ^2}x(2 - {\cos ^2}x) = {\sin ^2}x(1 + (1 - {\cos ^2}x)) \\
   = {\sin ^2}x(1 + {\sin ^2}x) - - - - (ii) \\
 $

From the equation (i) and (ii):

$
  {\sin ^2}x(2 - {\cos ^2}x) = {\sin ^2}x(1 + {\sin ^2}x) \\
   = \cos x(1 + \cos x) \\
   = \cos x + {\cos ^2}x \\
 $

Now, from the equation (a):
$
  {\sin ^2}x(2 - {\cos ^2}x) = \cos x + {\cos ^2}x \\
   = 1 \\
 $

Hence, ${\sin ^2}x(2 - {\cos ^2}x) = 1$.
Option A is correct.

Note:
Alternatively, the given trigonometric function can be reduced to a smaller expression by carrying out general algebraic and trigonometric identities. In general, this type of question can easily be solved by using trigonometric identities only.

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