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BabaBlast [244]
1 year ago
6

How do I verify this identity? I know you should write sin2a as sin(a+a).

Mathematics
1 answer:
Kisachek [45]1 year ago
7 0

SOLUTION

Given the question in the image, the following are the solution steps to verify the identity

STEP 1: Write the given identity

\sin 2\alpha=2\sin \alpha\cos \alpha

STEP 2: Verify the identity

\begin{gathered} \sin 2\alpha=2\sin \alpha\cos \alpha \\ \text{Consider the left hand side of the above trigonometry identity.} \\ \text{That is, }\sin 2\alpha\text{.} \\ \text{ Rewrite }\sin 2\alpha\text{ as }\sin (\alpha+a) \\ \text{ It is known that }\sin (a+b)=\sin (a)\cos b+\cos (a)\sin (b) \\ U\sin g\text{ this statement above, we have;} \\ \sin (\alpha+a)=\sin a\cos \alpha+\cos a\sin \alpha \\ \text{It is known that }xy+yx=xy+xy=2\times xy=2xy \\ U\sin g\text{ this statement above, we have;} \\ \sin a\cos \alpha+\cos a\sin \alpha=\sin a\cos \alpha+\sin \alpha\cos \alpha=2\times\sin \alpha\cos \alpha=2\sin \alpha\cos \alpha \\ \text{Hence, }\sin 2\alpha=2\sin \alpha\cos \alpha \end{gathered}

The verification of the identity is as seen above.

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Step-by-step explanation:

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Aleonysh [2.5K]
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3 years ago
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nikdorinn [45]

Answer:

Graph of the inequality 3y-2x>-18 is given below.

Step-by-step explanation:

We are given the inequality, 3y-2x>-18

Now, using the 'Zero Test', which states that,

After substituting the point (0,0) in the inequality, if the result is true, then the solution region is towards the origin. If the result is false, then the solution region is away from the origin'.

So, after substituting (0,0) in 3y-2x>-18, we get,  

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Thus, the solution region is towards the origin.

Hence, the graph of the inequality 3y-2x>-18 is given below.

8 0
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