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wel
3 years ago
13

In the figure, line m is || line n. Name a pair of supplementary angles.

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
8 0
Angle 1 and angle 2
OR
Angle 3 and Angle 4
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Derive these identities using the addition or subtraction formulas for sine or cosine: sinacosb=(sin(a+b)+sin(a-b))/2
Sergeu [11.5K]

Answer:

The work is in the explanation.

Step-by-step explanation:

The sine addition identity is:

\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).

The sine difference identity is:

\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a).

The cosine addition identity is:

\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b).

The cosine difference identity is:

\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b).

We need to find a way to put some or all of these together to get:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}.

So I do notice on the right hand side the \sin(a+b) and the \sin(a-b).

Let's start there then.

There is a plus sign in between them so let's add those together:

\sin(a+b)+\sin(a-b)

=[\sin(a+b)]+[\sin(a-b)]

=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]

There are two pairs of like terms. I will gather them together so you can see it more clearly:

=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]

=2\sin(a)\cos(b)+0

=2\sin(a)\cos(b)

So this implies:

\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)

Divide both sides by 2:

\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)

By the symmetric property we can write:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}

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The second option.

Basically, you can use SAS.

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inessss [21]

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Which will be the digit in ONE’S place in multiplication 3x5x7x9
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Answer:

5

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