trigonometry (1)

**thereddevils** · October 7th 2009, 12:31 AM

Show that $sin^2x + \frac{1}{2}cos x = \frac{17}{16}-(\cos x-\frac{1}{4})^2$

I can do this part .

Hence , state the range

$\frac{1}{2\sin^2 x+\cos x-1}$ for $0\leq x\leq 2\pi$

Need help with this part . Thanks .

**Grandad** · October 7th 2009, 12:43 AM

Hello thereddevils

Originally Posted by thereddevils

Show that $sin^2x + \frac{1}{2}cos x = \frac{17}{16}-(\cos x-\frac{1}{4})^2$

I can do this part .

Hence , state the range

$\frac{1}{2\sin^2 x+\cos x-1}$ for $0\leq x\leq 2\pi$

Need help with this part . Thanks .

$\frac{1}{2\sin^2 x+\cos x-1}= \frac{\frac12}{\sin^2x+\tfrac12\cos x - \tfrac12}$

$=\frac{\frac12}{\frac{17}{16}-(\cos x - \frac14)^2-\tfrac12}$

$=\frac{\frac12}{\frac{9}{16}-(\cos x - \frac14)^2}$

Now use the fact that $0\le(\cos x -\tfrac14)^2\le\tfrac{25}{16}$ to find the range of values of this expression. (Can you see why?)

Grandad

**thereddevils** · October 7th 2009, 12:49 AM

Originally Posted by Grandad

Hello thereddevils $\frac{1}{2\sin^2 x+\cos x-1}= \frac{\frac12}{\sin^2x+\tfrac12\cos x - \tfrac12}$

$=\frac{\frac12}{\frac{17}{16}-(\cos x - \frac14)^2-\tfrac12}$

$=\frac{\frac12}{\frac{9}{16}-(\cos x - \frac14)^2}$

Now use the fact that $0\le(\cos x -\tfrac14)^2\le\tfrac{25}{16}$ to find the range of values of this expression. (Can you see why?)

Grandad

Thanks Grandad , but i don see how you got that range ? Can you help me a little further

**Grandad** · October 7th 2009, 03:12 AM

Hello thereddevils

Originally Posted by thereddevils

Thanks Grandad , but i don see how you got that range ? Can you help me a little further

The cosine of any angle lies between $-1$ and $+1$ . So the maximum value of $(\cos x - \tfrac14)$ is $1-\tfrac14=\tfrac34$ and its minimum value is $-1-\tfrac14=-\tfrac54$ .

But when we square it, $(\cos x - \tfrac14)^2$ is always positive, and now lies between $0$ and $(-\tfrac54)^2 = \tfrac{25}{16}$ .

OK?

Grandad

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