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Question: If \[\cos \left( {\alpha + \beta } \right) = 0\], then \[\sin \left( {\alpha - \beta } \right)\] can...

If cos(α+β)=0\cos \left( {\alpha + \beta } \right) = 0, then sin(αβ)\sin \left( {\alpha - \beta } \right) can be reduced to?

Explanation

Solution

Here, we will write the RHS of the given equation in terms of cos function by using the trigonometric value. , find the value of α\alpha . Substituting the value of α\alpha in sin(αβ)\sin \left( {\alpha - \beta } \right) and solving it further using the trigonometric identities, we will get the required reduced value of sin(αβ)\sin \left( {\alpha - \beta } \right).

Formula Used:
We will use the formula sin(90θ)=cosθ\sin \left( {90^\circ - \theta } \right) = \cos \theta

Complete step-by-step answer:
It is given that cos(α+β)=0\cos \left( {\alpha + \beta } \right) = 0.
Now, we know that the value of cos90=0\cos 90^\circ = 0
Hence, substituting the value of 0 in the RHS of cos(α+β)=0\cos \left( {\alpha + \beta } \right) = 0, we get
cos(α+β)=cos90\Rightarrow \cos \left( {\alpha + \beta } \right) = \cos 90^\circ
Cancelling cos function from both sides, we get
(α+β)=90\Rightarrow \left( {\alpha + \beta } \right) = 90^\circ
Subtracting β\beta from both sides, we get
α=90β\alpha = 90^\circ - \beta …………………………. (1)\left( 1 \right)
Now, we have to reduce the value of sin(αβ)\sin \left( {\alpha - \beta } \right).
Now, substituting the value of α=90β\alpha = 90^\circ - \beta from equation (1)\left( 1 \right) into sin(αβ)\sin \left( {\alpha - \beta } \right), we get
sin(αβ)=sin(90ββ)\sin \left( {\alpha - \beta } \right) = \sin \left( {90^\circ - \beta - \beta } \right)
sin(αβ)=sin(902β)\Rightarrow \sin \left( {\alpha - \beta } \right) = \sin \left( {90^\circ - 2\beta } \right)
Now, using the formula sin(90θ)=cosθ\sin \left( {90^\circ - \theta } \right) = \cos \theta , we get
sin(αβ)=cos2β\Rightarrow \sin \left( {\alpha - \beta } \right) = \cos 2\beta
Hence, if cos(α+β)=0\cos \left( {\alpha + \beta } \right) = 0, then sin(αβ)\sin \left( {\alpha - \beta } \right)can be reduced to cos2β\cos 2\beta .
Therefore, this is the required answer.

Note: Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities are true for every value occurring on both sides of an equation. Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle. The six basic trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent. All these trigonometric ratios are defined using the sides of the right triangle, such as an adjacent side, opposite side, and hypotenuse side. All the fundamental trigonometric identities are derived from the six trigonometric ratios.