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Question

Question: Simplify: \(\sin \left( {\pi + \theta } \right)\sin \left( {\pi - \theta } \right)\cos e{c^2}\theta ...

Simplify: sin(π+θ)sin(πθ)cosec2θ\sin \left( {\pi + \theta } \right)\sin \left( {\pi - \theta } \right)\cos e{c^2}\theta .
(A) 11
(B) 1 - 1
(C) sinθ\sin \theta
(D) sinθ- \sin \theta

Explanation

Solution

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as sin(π+x)=sinx\sin \left( {\pi + x} \right) = - \sin x and sin(πx)=sinx\sin \left( {\pi - x} \right) = \sin x . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We will use the above mentioned trigonometric identities to simplify the expression and then open the brackets to get to the required answer.

Complete answer:
In the given problem, we have to simplify the product sin(π+θ)sin(πθ)cosec2θ\sin \left( {\pi + \theta } \right)\sin \left( {\pi - \theta } \right)\cos e{c^2}\theta .
So, sin(π+θ)sin(πθ)cosec2θ\sin \left( {\pi + \theta } \right)\sin \left( {\pi - \theta } \right)\cos e{c^2}\theta
We know the trigonometric formula sin(π+x)=sinx\sin \left( {\pi + x} \right) = - \sin x. So, we get,
== sinθsin(πθ)cosec2θ- \sin \theta \sin \left( {\pi - \theta } \right)\cos e{c^2}\theta
We also know the trigonometric formula sin(πx)=sinx\sin \left( {\pi - x} \right) = \sin x. So, we get,
== sinθsinθcosec2θ- \sin \theta \sin \theta \cos e{c^2}\theta
Now, expressing the product of sinθ\sin \theta with itself as sin2θ{\sin ^2}\theta , we get,
== sin2θcosec2θ- {\sin ^2}\theta \cos e{c^2}\theta
Now, we know that cosecant and sine are reciprocal trigonometric functions of each other. So, we get the expression as,
== sin2θ  sin2θ - \dfrac{{{{\sin }^2}\theta }}{{\;{{\sin }^2}\theta }}
Now, cancelling common factors in numerator and denominator, we get,
== 1 - 1
Hence, the product sin(π+θ)sin(πθ)cosec2θ\sin \left( {\pi + \theta } \right)\sin \left( {\pi - \theta } \right)\cos e{c^2}\theta can be simplified as (1)( - 1) by the use of basic algebraic rules and simple trigonometric formulae.
So, option (B) is the correct answer.

Additional information: Trigonometric functions are also called Circular functions. Trigonometric functions are the functions that relate an angle of a right angled triangle to the ratio of two side lengths. There are 66 trigonometric functions, namely: sin(x)\sin (x),cos(x)\cos (x),tan(x)\tan (x),cosec(x)\cos ec(x),sec(x)\sec (x) and cot(x)\cot \left( x \right) . Also, cosec(x)\cos ec(x) ,sec(x)\sec (x)and cot(x)\cot \left( x \right) are the reciprocals of sin(x)\sin (x),cos(x)\cos (x) and tan(x)\tan (x) respectively.

Note:
The problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as: sin(π+x)=sinx\sin \left( {\pi + x} \right) = - \sin x and sin(πx)=sinx\sin \left( {\pi - x} \right) = \sin x . Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. We can also simplify the given expression using the compound angle formulae for sine. Questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers.