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adell [148]
3 years ago
13

Question 14 (5 points)

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
7 0

Answer:

14) C

15) D, because 90:20=4.5

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Step-by-step explanation:

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Hello! Verify the identity. Please show your work! Use trigonometric identities to verify each expression is equal.
RSB [31]

Answer:

See Below.

Step-by-step explanation:

We want to verify the identity:

\displaystyle \csc^2 x -2\csc x \cot x +\cot ^2 x = \tan^2\left(\frac{x}{2}\right)

Note that the left-hand side is a perfect square trinomial pattern. Namely:

a^2-2ab+b^2=(a-b)^2

If we let <em>a</em> = csc(x) and <em>b</em> = cot(x), we can factor it as such:

\displaystyle (\csc x - \cot x)^2 = \tan^2\left(\frac{x}{2}\right)

Let csc(x) = 1 / sin(x) and cot(x) = cos(x) / sin(x):

\displaystyle \left(\frac{1}{\sin x}-\frac{\cos x }{\sin x}\right)^2=\tan^2\left(\frac{x}{2}\right)

Combine fractions:

\displaystyle \left(\frac{1-\cos x}{\sin x}\right)^2=\tan^2\left(\frac{x}{2}\right)

Square (but do not simplify yet):

\displaystyle \frac{(1-\cos x)^2}{\sin ^2x}=\tan^2\left(\frac{x}{2}\right)

Now, we can make a substitution. Let <em>u</em> = <em>x</em> / 2. So, <em>x</em> = 2<em>u</em>. Substitute:

\displaystyle \frac{(1-\cos 2u)^2}{\sin ^22u}=\tan^2u

Recall that cos(2u) = 1 - sin²(u). Hence:

\displaystyle \frac{(1-(1-2\sin^2u))^2}{\sin ^2 2u}=\tan^2u

Simplify:

\displaystyle \frac{4\sin^4 u}{\sin ^2 2u}=\tan^2 u

Recall that sin(2u) = 2sin(u)cos(u). Hence:

\displaystyle \frac{4\sin^4 u}{(2\sin u\cos u)^2}=\tan^2 u

Square:

\displaystyle \frac{4\sin^4 u}{4\sin^2 u\cos ^2u}=\tan^2 u

Cancel:

\displaystyle \frac{\sin ^2 u}{\cos ^2 u}=\tan ^2 u

Since sin(u) / cos(u) = tan(u):

\displaystyle \left(\frac{\sin u}{\cos u}\right)^2=\tan^2u=\tan^2u

We can substitute <em>u</em> back for <em>x</em> / 2:

\displaystyle \tan^2\left(\frac{x}{2}\right)= \tan^2\left(\frac{x}{2}\right)

Hence proven.

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