MHB Find cos8A Using cos2A=sqrt(m)

[NotaMathPerson]

If cos2A=sqrt(m) find cos8A

I used cos2A = (cosA)^2-(sinA)^2

Please help me continue

 

[MarkFL]

Instead of a double-angle identity for cosine, try the half-angle identity:

$$\cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{1+\cos(\theta)}{2}}$$

So, what you get is:

$$\frac{1+\cos(4A)}{2}=m$$

Can you continue?

 

[NotaMathPerson]

I get 8m^2-8+1.

- - - Updated - - -

If sinA=4/5, A isin quadrant 2, sinB=7/25 B is in quadrant 1. Find cos (A-B)

What I did is find A and B by getting the arcsin of both sinA and sinB

The get the value for cos(A-B)

is my method correct. Please help. Thanks.

 

[Greg]

Hi NotaMathPerson. In future, please start a new thread for each new problem. This helps prevent the original thread from becoming convoluted. :)

For your first problem I get $8m^2-8m+1$.

For your second problem, can you calculate the values of cos(A) and cos(B) and then use the angle sum and difference formula for cosine:

cos(A - B) = cos(A)cos(B) + sin(A)sin(B)

?

 

[NotaMathPerson]

Hello!

I get cos(A-B)=(3/5)(24/25)+(4/5)(7/25)= 0.8 is thus correct?

 

[Greg]

As angle A is quadrant 2, cos(A) = -3/5.

 

[NotaMathPerson]

As angle A is quadrant 2, cos(A) = -3/5.

Oh yes. I forgot.

It should be -44/125

 

[soroban]

\text{If }\,\cos2A\,=\,\sqrt{m},\,\text{ find }\cos8A.

\text{I used: }\,\cos2A =\, \,\cos^2A -\sin^2A

Please help me continue.

Instead, use: \cos2A \:=\:2\cos^2\!A - 1

\begin{array}{cccccc}\text{Then:} &amp; \cos4A &amp;=&amp; 2\cos^2\!2A-1 \\ \\<br /> <br /> \text{And:} &amp; \cos8A &amp; =&amp; 2\cos^2\!4A-1 \\ \\ <br /> <br /> &amp;&amp; = &amp; 2(2\cos^2\!2A - 1)^2 - 1 \\ \\<br /> <br /> &amp;&amp; = &amp; 2(4\cos^4\!2A - 4\cos^2\!2A + 1) - 1 \\ \\<br /> <br /> &amp;&amp; = &amp; 8\cos^4\!2A - 8\cos^2\!2A + 2 - 1 \\ \\<br /> <br /> &amp;&amp; = &amp; 8\cos^42A - 8\cos^22A + 1 \\ \\<br /> <br /> &amp;&amp; = &amp; 8(\sqrt{m})^4 - 8(\sqrt{m})^2 + 1 \\ \\<br /> <br /> &amp;&amp; = &amp; 8m^2 - 8m + 1<br /> \end{array}