Question

HELP ME PLEASE .
1.use a half-angle formula to find the exact value of sin 67.5°.
4. find the exact solution in the interval [0,2pi] for tanx/2=sinx/2
5. rewrite the following expression in terms of the first power if cosine tan^2xsin^2x

Answer 1

Since we know that: \(\cos(2x)=1-2\sin^2(x)\)
We can also say that:
\[\cos(\phi)=1-2\sin^2\left(\frac{\phi}{2}\right)\]
And we can isolate for sin:
\[\sin\frac{\phi}{2}=\pm\sqrt{\frac{1-cos(\phi)}{2}}\]
So then we are given the angle 67.5 degrees. Notice how \(67.5=\frac{135}{2}\)
So therefore:
\[\sin\frac{135}{2}=\pm\sqrt{\frac{1-\cos135}{2}}\]

Answer 2

Im not done yet haha

Answer 3

You can find the exact value of \(cos135\) by using a 45-45-90 triangle. It's real easy if you haven't already learned it:

So then If we know that the \(\cos135^\circ=\frac{1}{\sqrt{2}}\) Then we can simplify the answer even more. Just give me a sec

Answer 4

thanks

Answer 5

K so then:
\[\sin\frac{135}{2}=\pm\sqrt{\frac{1-\frac{1}{\sqrt2}}{2}}=\pm\sqrt{\frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{2}}=\pm\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}\]