trig value help

Tweety · #1 $Old$ November 17th, 2009, 04:21 AM

Calculate without using a calculator the exact value of;

$\sqrt{3}cos15-sin15$

I dont know which trig identity to use in the first place, how would I know? And I assume I meant to be using the special triangles, since I am not meant to use a calculator?

mr fantastic · #2 $Old$ November 17th, 2009, 04:28 AM

Quote:

Originally Posted by Tweety $View Post$

Calculate without using a calculator the exact value of;

$\sqrt{3}cos15-sin15$

I dont know which trig identity to use in the first place, how would I know? And I assume I meant to be using the special triangles, since I am not meant to use a calculator?

Note that 15 = 45 - 30 (and both 45 and 30 are, in degrees, special angles).

Tweety · #3 $Old$ November 17th, 2009, 04:33 AM

Quote:

Originally Posted by mr fantastic $View Post$

Note that 15 = 45 - 30 (and both 45 and 30 are, in degrees, special angles).

Thanks, so would it be something like this;

$\sqrt{3}cos(45-30)-sin(45-30)$ ?

than expand using sin(A-B) AND cos(A-B)

mr fantastic · #4 $Old$ November 17th, 2009, 04:36 AM

Quote:

Originally Posted by Tweety $View Post$

Thanks, so would it be something like this;

$\sqrt{3}cos(45-30)-sin(45-30)$ ?

than expand using sin(A-B) AND cos(A-B)

Yes, and then use the compound angle formula.

Tweety · #5 $Old$ November 17th, 2009, 05:00 AM

Quote:

Originally Posted by mr fantastic $View Post$

Yes, and then use the compound angle formula.

Okay thanks,
But I don't know what the 'compound angle' formula is, (?) we just started doing trig addition formula's in class so maybe we have not got up to it yet.

After simplyfying I got $\frac{\sqrt{28}}{2}$

and my books says the right answer is $\sqrt{2}$

mathaddict · #6 $Old$ November 17th, 2009, 06:36 AM

Quote:

Originally Posted by Tweety $View Post$

Okay thanks,
But I don't know what the 'compound angle' formula is, (?) we just started doing trig addition formula's in class so maybe we have not got up to it yet.

After simplyfying I got $\frac{\sqrt{28}}{2}$

and my books says the right answer is $\sqrt{2}$

HI

Try spot your mistake here .

$\sqrt{3}[\cos 45\cos 30+\sin 45\sin 30]-\sin 45\cos 30+\cos 45\sin 30$

$=\sqrt{3}[\frac{1}{\sqrt{2}}(\frac{\sqrt{3}}{2})+\frac{1}{\sqrt{2}}(\frac{1}{2})]-\frac{1}{\sqrt{2}}(\frac{\sqrt{3}}{2})+\frac{1}{\sqrt{2}}(\frac{1}{2})$

Them simplify and rationalise the surds to get the answer .

Tweety · #7 $Old$ November 17th, 2009, 07:27 AM

Quote:

Originally Posted by mathaddict $View Post$

HI

Try spot your mistake here .

$\sqrt{3}[\cos 45\cos 30+\sin 45\sin 30]-\sin 45\cos 30+\cos 45\sin 30$

$=\sqrt{3}[\frac{1}{\sqrt{2}}(\frac{\sqrt{3}}{2})+\frac{1}{\sqrt{2}}(\frac{1}{2})]-\frac{1}{\sqrt{2}}(\frac{\sqrt{3}}{2})+\frac{1}{\sqrt{2}}(\frac{1}{2})$

Them simplify and rationalise the surds to get the answer .

Thank you,

Is there not meant to be a minus sigh in front of the last part? because its sin(A-B)= sinAcosB - cosAsinB

Grandad · #8 $Old$ November 17th, 2009, 08:00 AM

Hello Tweety

Quote:

Originally Posted by Tweety $View Post$

Calculate without using a calculator the exact value of;

$\sqrt{3}cos15-sin15$

I dont know which trig identity to use in the first place, how would I know? And I assume I meant to be using the special triangles, since I am not meant to use a calculator?

I realise that you've done lots of work on this question already, so you might not be interested in a different method. But here's one that avoids all the hard work, by using the following:

$\cos 30^o=\frac{\sqrt3}{2}, \sin 30^o = \frac12 \text{ and} \cos45^o=\frac{1}{\sqrt2}$

(and if you didn't know those facts, then learn them!)

$\sqrt3\cos15^o -\sin15^o = 2\left(\frac{\sqrt3}{2}\cos15^o-\frac{1}{2}\sin 15^o\right)$

$=2(\cos30^o\cos15^o-\sin30^o\sin15^o)$

$=2\cos(30^o+15^o)$

$=2\cos45^o$

$=\frac{2}{\sqrt2}$

$=\sqrt2$

I suspect that was the method that the person who set the question was hoping you'd find!

Grandad

pacman · #9 $Old$ November 18th, 2009, 07:12 AM

thanks Grandad, i like your method. It is self-instructive.