+108 Hi Forum!
Here are the questions.
1. There is an acute angle theta such that cos(theta)=1/2+sin(theta). Find cos(2theta).
2. If the angle theta is such that 2pi is less than or equal to theta which is less than or equal to 4pi and cos(theta) = -7/25, then what is the value of sin(theta/2)?
Fast replies is needed if at all possible. Thanks again! - BasicMaths
+118609 1. There is an acute angle theta such that cos(theta)=1/2+sin(theta). Find cos(2theta).
\(cos(\theta)=1/2+sin(\theta)\\ find\; \;\;cos(2\theta)\\~\ cos(2\theta)=cos^2\theta-\sin^2\theta\\ so\\ cos(2\theta)=(0.5+sin\theta)^2-\sin^2\theta\\ \)
In the interest of not promoting cheating. can you take it from there?
+108 Ok, so let me work this out:
If I square the stuff inside parentheses I get 0.25 + sin(theta) + sin^2(theta)-sin^2(theta).
The sin squares cancel, giving us 0.25+sin(theta). This is good so far, but this is where I got stuck last time.........Help!
+118609 1. continued
\(cos(2\theta)=(0.5+sin\theta)^2-sin^2\theta\\ cos(2\theta)=(0.25+sin\theta+sin^2\theta)-sin^2\theta\\ cos(2\theta)=0.25+sin\theta\\ or\\ cos(2\theta)=0.25+cos\theta-0.5\\ cos(2\theta)=cos\theta-0.25\\ \)
\(cos2\theta=cos^2\theta-sin^2\theta\\ cos2\theta=1-2sin^2\theta\\ so\\ 1-2sin^2\theta=0.25+sin\theta\\ 2sin^2\theta+sin\theta-0.75=0\\ let\;\; x=sin\theta\\ 2x^2+x-0.75=0\\ x=\frac{-1\pm\sqrt{1+6}}{4}\\ x=\frac{-1\pm\sqrt{7}}{4}\\\)
x must be positive so
\(sin\theta = \frac{\sqrt7 -1 }{4}\\ sin^2\theta=\frac{7+1-2\sqrt7}{16}\\ sin^2\theta=\frac{8-2\sqrt7}{16}\\ cos^2\theta=1-\frac{8-2\sqrt7}{16}\\ sin^2\theta-cos^2\theta = \frac{8-2\sqrt7}{16}-(1-\frac{8-2\sqrt7}{16})\\ sin^2\theta-cos^2\theta = 2*\frac{8-2\sqrt7}{16}-1\\ sin^2\theta-cos^2\theta = \frac{4-\sqrt7}{4}-1\\ sin^2\theta-cos^2\theta = \frac{-\sqrt7}{4}\\ cos(2\theta)= \frac{-\sqrt7}{4}\\\)
I have not checked this AT ALL. So you better check carefully for stupid mistakes.
+118609 You need to do the 2nd one yourself. Or at least show a solid attempt before anyone should help you.
