The result of expanding the trigonometry expression $\sin^2(\theta) * (1 + \cos(\theta))$ is $cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

<h3>How to evaluate the expression?</h3>

The expression is given as:

$\sin^2(\theta) * (1 + \cos(\theta))$

Express $\sin^2(\theta)$ as $1 - \cos^2(\theta)$ .

So, we have:

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

Open the bracket

$\sin^2(\theta) * (1 + \cos(\theta)) = 1 + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

Express 1 as cos°(Ф)

$\sin^2(\theta) * (1 + \cos(\theta)) = cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

Hence, the result of expanding the trigonometry expression $\sin^2(\theta) * (1 + \cos(\theta))$ is $cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

Read more about trigonometry expressions at:

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3 0

2 years ago

Shellie spent $10.56 on oranges that cost $0.88 per orange. How many oranges did she purchase?

vlada-n [284]

You have to do 10.56 plus 88 and you will have your answer

6 0

3 years ago

The proof that AMNS AQNS is shown.

FrozenT [24]

Answer:

The answer is "MS and QS".

Step-by-step explanation:

Given ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. we have to prove that ΔMNS ≅ ΔQNS.

As NR and MQ bisect each other at S

⇒ segments MS and SQ are therefore congruent by the definition of bisector i.e MS=SQ

In ΔMNS and ΔQNS

MN=QN (∵ MNQ is isosceles triangle)

∠NMS=∠NQS (∵ MNQ is isosceles triangle)

MS=SQ (Given)

By SAS rule, ΔMNS ≅ ΔQNS.

Hence, segments MS and SQ are therefore congruent by the definition of bisector.

The correct option is MS and QS

3 0

3 years ago