Question

Prove that sin (240+theta)+cos(330+theta)=0

Answer 1

Make use of the following identities:

$\sin (180^o+\theta)=-\sin (\theta)\\ \cos (360^o \pm \theta)=\cos (\theta)\\ \cos (90^o-\theta)=\sin (\theta)$

Using the above identities,

$\cos (330^o + \theta)=\cos (360^o + \theta-30^o)=\cos ( \theta-30^o)=\sin(60^o+\theta)$

$\sin (240^o+\theta)=-\sin(60^o+\theta)$

$\sin (180^o+\theta)+\cos (330^o + \theta)=-\sin(60^o+\theta)+\sin(60^o+\theta)=0=RHS$

The proof is complete.

Answer 2

Step-by-step explanation:

$\sin(240 + theta) + \cos(330 + theta) \\ = \sin(240) \cos(theta) + \cos(240) \sin(theta) + \cos(330) \cos(theta) - \sin(330) \sin(theta) \\ = ( \sin(270 - 30) ) \cos(theta) + (\cos(270 - 30) ) \sin(theta) + (\cos(360 - 30) ) \cos(theta) - (\sin(360 - 30) ) \sin(theta) \\ = ( \sin(270) \cos(30) - \cos(270) \sin(30) ) \cos(theta) + ( \cos(270) \cos(30) + \sin(270) \sin(30) ) \sin(theta) + ( \cos(360) \cos(30) + \sin(360) \sin(30) ) \cos(theta) - ( \sin(360)\cos(30) - \cos(360) \sin(30)) \sin(theta) \\ = ( - 1 \times \frac{ \sqrt{3} }{2} - 0 \times \frac{1}{2} ) \cos(theta) + (0 \times \frac{ \sqrt{3} }{2} + 1 \times \frac{1}{2} ) \sin(theta) + (1 \times \frac{ \sqrt{3} }{2} + 0 \times \frac{1}{2} ) \cos(theta) - (0 \times \frac{ \sqrt{3} }{2} - 1 \times \frac{1}{2} ) \sin(theta) \\ = - \frac{ \sqrt{3} }{2} \cos(theta) + \frac{1}{2} \sin(theta) + \frac{ \sqrt{3} }{2} \cos(theta) - ( - \frac{1}{2} \sin(theta)) \\ = - \frac{ \sqrt{3} }{2} \cos(theta) + \frac{ \sqrt{3} }{2} \cos(theta) + \frac{1}{2} \sin(theta) + \frac{1}{2} \sin(theta) \\ = 0 + \sin(theta) = \sin(theta)$

since

$\sin(theta) \: is \: not = 0$

Hence proved

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