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11111nata11111 [884]
3 years ago
6

Solve trigonometric function sin∅ × csc∅ + cos(3π/2 - ∅) × sin∅

Mathematics
2 answers:
hoa [83]3 years ago
6 0

Answer:

\sf \cos ^2\left(x\right)

Explanation:

\sf \sin \left(x\right)\csc \left(x\right)+\cos \left(\frac{3\pi }{2}-x\right)\sin \left(x\right)

\sf \sin \left(x\right)\csc \left(x\right)+\left(-\sin \left(x\right)\right)\sin \left(x\right)

\sf \sin \left(x\right)\csc \left(x\right)-\sin ^2\left(x\right)

\sf \cos ^2\left(x\right)

lapo4ka [179]3 years ago
4 0

Answer:

\cos^2(\theta)

Step-by-step explanation:

<u>Identities used:</u>

\csc(\theta)=\dfrac{1}{\sin(\theta)}

\cos(\frac{3 \pi}{2}-\theta)=\cos(\frac{3 \pi}{2})\cos(\theta)+\sin(\frac{3 \pi}{2})\sin(\theta)

\textsf{As }\cos(\frac{3 \pi}{2})=0\textsf{ and }\sin(\frac{3 \pi}{2})=-1

\implies \cos(\frac{3 \pi}{2}-\theta)=0 \times\cos(\theta)+-1\times\sin(\theta)=-\sin(\theta)

\sin^2(\theta)+\cos^2(\theta)=1 \implies \cos^2(\theta)=1-\sin^2(\theta)

Therefore,

\sin(\theta) \times \csc(\theta)+\cos(\frac{3 \pi}{2}-\theta)\times\sin(\theta)

=\sin(\theta) \times\dfrac{1}{\sin(\theta)}-\sin(\theta)\times\sin(\theta)

=1-\sin^2(\theta)

= \cos^2(\theta)

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