Question

cos theta + sin (270+ theta) - sin (270- theta) + cos ( 180+ theta ) = 0​

Answer 1

Step-by-step explanation:

LHS=cosθ+sin(270+θ)-sin(270-θ)+cos(180+θ)

=cosθ-cosθ+cosθ-cosθ (because sin(270+θ)=-cosθ,-sin(270-θ)=cosθ,cos(180+θ)=-cosθ)

=0

=RHS

ask if you have any questions:)

Answer 2

$\cos \theta + \sin (270+ \theta) - \sin (270- \theta) + \cos ( 180+ \theta ) = 0$ Proved.

Step-by-step explanation:

Consider the provided information.

$\cos \theta + \sin (270+ \theta) - \sin (270- \theta) + \cos ( 180+ \theta ) = 0$

Consider the Left hand side of the equation.

$\cos \theta+\sin[180+(90+\theta)]-\sin[180+(90- \theta)] + \cos ( 180+ \theta )$

Use the identity: $\sin(180+\theta)=-\sin\theta,\cos(180+\theta)=-\cos\theta$

$\cos \theta-\sin(90+\theta)+\sin(90- \theta) -\cos\theta$

Use the identity: $\sin(90+\theta)=\cos\theta,\sin(90-\theta)=\cos\theta$

$\cos \theta-\cos\theta+\cos\theta -\cos\theta$

$0$

LHS = RHS

Hence, proved

#Learn more

Prove sin(90°-x) = cos x.

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