Home › Forums › Math questions › Trigonometry questions

Find the value of

Posted July 8th, 2009 by alpha
in

If $ cos x -sinx=\frac{1}{2} $
then find the value of 1)cox+sinx
2)cos x
3)sinx

‹ Identity Find the value of ›

Comments

answer

On July 9th, 2009 Structure says:
$$\cos x-\sin x=\sqrt2\cos(x+\frac{\pi}{4})=\sqrt2(\frac{1}{\sqrt2}\cos x-\frac{1}{\sqrt2}\sin x)=\sqrt2(\cos\frac{\pi}{4}\cos x-\sin \frac{\pi}{4}\sin x)=\sqrt2\cos(x+\frac{\pi}{4})=\frac{1}{2}$$
$$\cos x+\sin x=\sqrt2(\frac{1}{\sqrt2}\cos x+\frac{1}{\sqrt2}\sin x)=\sqrt2(\sin\frac{\pi}{4}\cos x+\cos \frac{\pi}{4}\sin x)=\sqrt2\sin(x+\frac{\pi}{4})=\pm \sqrt2\sqrt{1-\frac{1}{8}}=\pm\frac{\sqrt 7}{2}$$

answer

On July 9th, 2009 Structure says:
$$\cos x-\sin x=\sqrt2\cos(x+\frac{\pi}{4})=\frac{1}{2}$$
$$\cos x+\sin x=\sqrt2(\frac{1}{\sqrt2}\cos x+\frac{1}{\sqrt2}\sin x)=\sqrt2(\sin\frac{\pi}{4}\cos x+\cos \frac{\pi}{4}\sin x)=\sqrt2\sin(x+\frac{\pi}{4})=\pm \sqrt2\sqrt{1-\frac{1}{8}}=\pm\frac{\sqrt 7}{2}$$

I don't follow your first step.

On July 9th, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

~~THANK YOU ~~

On July 10th, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Back to top