Solveeit Logo
Login

Question

Question: How do you prove \(\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \left( \theta \right)?\)...

How do you prove cos⁡(π2+θ)=−sin⁡(θ)?\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \left( \theta \right)?

Explanation

Solution

To prove the given equation, start evaluating the expression of the left hand side and try to make it equal to the expression at right hand side with the help of compound angle formula for cosine.

Formula used:
Compound angle formula for cosine addition: cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos (a + b) = \cos a\cos b - \sin a\sin b

Complete step by step answer:
In order to prove the given trigonometric equation, we will simplify the left hand side expression and make it equal to the right hand side expression of the equation.So in the equation,
cos⁡(π2+θ)=−sin⁡(θ)\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \left( \theta \right)
L.H.S.=cos⁡(π2+θ){\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \cos \left( {\dfrac{\pi }{2} + \theta } \right)
Now simplifying it with the help of compound angle formula for cosine, that is given as
cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos (a + b) = \cos a\cos b - \sin a\sin b
Simplifying left hand side expression with this formula as follows
cos⁡(π2+θ)=cos⁡π2cos⁡θ−sin⁡π2sin⁡θ\cos \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \dfrac{\pi }{2}\cos \theta - \sin \dfrac{\pi }{2}\sin \theta
Now from the basic trigonometric values table, we know the values of sine and cosine of the argument π2\dfrac{\pi }{2} that is 1  and 01\;{\text{and}}\,0 respectively. It can also be written as sin⁡π2=1  and  cos⁡π2=0\sin \dfrac{\pi }{2} = 1\;{\text{and}}\;\cos \dfrac{\pi }{2} = 0
Substituting these values in above expression, we will get

⇒cos⁡π2cos⁡θ−sin⁡π2sin⁡θ=0−sin⁡θ ⇒cos⁡π2cos⁡θ−sin⁡π2sin⁡θ=−sin⁡θ ∴cos⁡π2cos⁡θ−sin⁡π2sin⁡θ=R.H.S\Rightarrow\cos \dfrac{\pi }{2}\cos \theta - \sin \dfrac{\pi }{2}\sin \theta = 0 - \sin \theta \\\ \Rightarrow\cos \dfrac{\pi }{2}\cos \theta - \sin \dfrac{\pi }{2}\sin \theta = - \sin \theta \\\ \therefore\cos \dfrac{\pi }{2}\cos \theta - \sin \dfrac{\pi }{2}\sin \theta = {\text{R}}{\text{.H}}{\text{.S}}

Since we have simplified the expression at left hand side and made it equal to the expression at right hand side, hence proved the given trigonometric equation.

Note: The compound angle formulas of cosine for addition and subtraction have a bit difference and also a bit confusing, so remember the difference between them with a trick, if the formula for cosine difference is wanted then addition is there, and if for cosine addition is wanted then subtraction is there.This question can be proved with the help of basic trigonometric definitions and quadrants in graph, try to prove it with this method by yourself.