Question

What is the value of $\bf{sin^{2}\theta}$?​

Answer 1

Answer:

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What is the value of $\bf{sin^{2}\theta}$?

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Solution:

Given,

→ 5 tan θ = 4

• We have to find the value of (5 sin θ - 3 cos θ)/(5 sin θ + 2 cos θ)

So,

→ 5 tan θ = 4

→ tan θ = 4/5

Now,

\rm \dfrac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta}5sinθ+2cosθ5sinθ−3cosθ

Dividing both sides by cos θ, we get,

\rm = \dfrac{ \dfrac{5 \sin \theta - 3 \cos \theta}{ \cos \theta }}{ \dfrac{5 \sin \theta + 2 \cos \theta}{ \cos \theta }}=cosθ5sinθ+2cosθcosθ5sinθ−3cosθ

\rm = \dfrac{5 \dfrac{ \sin \theta}{ \cos \theta } - 3}{ 5\dfrac{ \sin \theta}{ \cos \theta } + 2}=5cosθsinθ+25cosθsinθ−3

We know that,

\sf \implies \dfrac{ \sin \theta}{ \cos \theta } = \tan \theta⟹cosθsinθ=tanθ

So,

\rm \dfrac{5 \dfrac{ \sin \theta}{ \cos \theta } - 3}{ 5\dfrac{ \sin \theta}{ \cos \theta } + 2}5cosθsinθ+25cosθsinθ−3

\rm = \dfrac{5 \tan \theta - 3}{ 5\tan \theta + 2}=5tanθ+25tanθ−3

\rm = \dfrac{5 \times \dfrac{4}{5} - 3}{ 5 \times \dfrac{4}{5} + 2}=5×54+25×54−3

\rm = \dfrac{4 - 3}{4+ 2}=4+24−3

\rm = \dfrac{1}{6}=61

So,

\rm \implies \dfrac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} = \dfrac{1}{6}⟹5sinθ+2cosθ5sinθ−3cosθ=61

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Answer 2

Answer:

The sin of 2 radians is 0.9093, the same as sin of 2 radians in degrees. To change 2 radians to degrees multiply 2 by 180° / = 114.59156°. Sin 2 = sin 114.59156 degrees.

Step-by-step explanation:

Hope it will help you..