2 Trig Questions

**Blahdkm** · Jun 18th 2009, 03:01 AM

A) I have to find the exact value of cos8*cos38* + sin8*sin38*(the *'s are degrees)

Now, I'm 90% sure I have to use cos(A+/-B)=cosAcosB +/- sinAsinB, but I've never understood what to do from that point.

B) I also have to prove that sin^2(x)/1-cos(x)=sec(x)+1/sec(x), yet I've tried several times and reach nothing but deadends.

Mucho Gracias for the help in advance!

**yeongil** · Jun 18th 2009, 03:17 AM

Originally Posted by Blahdkm

A) I have to find the exact value of cos8*cos38* + sin8*sin38*(the *'s are degrees)

Now, I'm 90% sure I have to use cos(A+/-B)=cosAcosB +/- sinAsinB, but I've never understood what to do from that point.

$\cos (A - B) = \cos A \cos B + \sin A \sin B$ , so

$\cos 8^{\circ} \cos 38^{\circ} + \sin 8^{\circ} \sin 38^{\circ} = \cos (8^{\circ} - 38^{\circ}) = \cos (-30^{\circ}) = \frac{\sqrt{3}}{2}$

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**yeongil** · Jun 18th 2009, 03:22 AM

Originally Posted by Blahdkm

B) I also have to prove that sin^2(x)/1-cos(x)=sec(x)+1/sec(x), yet I've tried several times and reach nothing but deadends.

You mean this?
$\frac{\sin^2 x}{1 - \cos x} = \frac{\sec x + 1}{\sec x}$ ?

$\frac{\sin^2 x}{1 - \cos x}$
$\begin{aligned} &= \frac{1 - \cos^2 x}{1 - \cos x} \\ &= \frac{(1 + \cos x)(1 - \cos x)}{1 - \cos x} \\ &= 1 + \cos x \end{aligned}$
$\begin{aligned} &= 1 + \frac{1}{\sec x} \\ &= \frac{\sec x}{\sec x} + \frac{1}{\sec x} \\ &= \frac{\sec x + 1}{\sec x} \end{aligned}$

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