# trigonometry (1)

• Oct 7th 2009, 12:31 AM
thereddevils
trigonometry (1)
Show that $\displaystyle sin^2x + \frac{1}{2}cos x = \frac{17}{16}-(\cos x-\frac{1}{4})^2$

I can do this part .

Hence , state the range

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

Need help with this part . Thanks .
• Oct 7th 2009, 12:43 AM
Hello thereddevils
Quote:

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

I can do this part .

Hence , state the range

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

Need help with this part . Thanks .

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

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

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

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

• Oct 7th 2009, 12:49 AM
thereddevils
Quote:

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

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

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

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

Thanks Grandad , but i don see how you got that range ? Can you help me a little further (Rofl)
• Oct 7th 2009, 03:12 AM
The cosine of any angle lies between $\displaystyle -1$ and $\displaystyle +1$. So the maximum value of $\displaystyle (\cos x - \tfrac14)$ is $\displaystyle 1-\tfrac14=\tfrac34$ and its minimum value is $\displaystyle -1-\tfrac14=-\tfrac54$.
But when we square it, $\displaystyle (\cos x - \tfrac14)^2$ is always positive, and now lies between $\displaystyle 0$ and $\displaystyle (-\tfrac54)^2 = \tfrac{25}{16}$.