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Given that:
$\tan A=1$ and $\tan B = \sqrt{3}$

How would you find $\cos A \cos B - \sin A \sin B$?


EDIT: This is what I've tried after reading bhattacharjee's answer:

$$ \tan(A+B) = \tan A+\tan B−\tan A\tan B$$ so, $\tan(A+B)= {1+\sqrt{3} \over 1-\sqrt{3}}$

from this I get $1 \over \cos(A+B)^2 $ $=1+ \left ( {1+\sqrt{3} \over 1-\sqrt{3}}\right )^2$

=> $ 1 \over \cos^2(A+B) $ $=$ $ 8 \over 4-2 \sqrt{3}$

Is this right, because it seems like a dead end to me? How am I supposed to proceed from here?

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  • $\begingroup$ From here, flip the fractions and take the square root... $\endgroup$ – The Chaz 2.0 Apr 20 '13 at 17:57
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HINT:

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

and $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$

Do you know how to find $\cos \theta$ from $\tan\theta?$

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  • $\begingroup$ No i dont, please provide the entire solution. $\endgroup$ – Ghost Apr 20 '13 at 16:56
  • $\begingroup$ @Ghost, $\cos\theta=\frac1{\sec\theta}$ and $\sec^2\theta=1+\tan^2\theta$. $\endgroup$ – lab bhattacharjee Apr 20 '13 at 16:58
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    $\begingroup$ Yeah, well..."entire solutions" are not the best way to learn. You can either check your notes/book(s) or even one of the several millions of sites where this stuff is treated... $\endgroup$ – DonAntonio Apr 20 '13 at 16:59
  • $\begingroup$ Hmmm thanks, I've tried solving it once again. But I've hit an apparent dead end. Care for a little help? $\endgroup$ – Ghost Apr 20 '13 at 17:30
  • $\begingroup$ @Ghost, So, $$\cos^2(A+B)=\frac{4-2\sqrt3}8=\frac{(\sqrt3-1)^2}8\implies \cos(A+B)=\pm\frac{\sqrt3-1}{2\sqrt2}$$ $\endgroup$ – lab bhattacharjee Apr 20 '13 at 17:52
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It is given that tanA=1, and tan B=$/sqrt3$.

This implies that A can be 45 degrees and B can be 60 degrees.( Taking inverse, and using principal values)

Also cosA.cosB-sinAsinB=cos(A B)

So, cos(A B)=cos(60 45)=cos(105)=-sin(15)=-($\sqrt3$-1)/(2$\sqrt2$)

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since $tan A =1$, then $A = 45^0$ and since $tan B = \sqrt{3}$, then $ B = 60^0$.

By drawing triangles you can find that $sin 45^0 = 1/\sqrt{2}$ and $cos 45^0 = 1/\sqrt{2}$ and $sin 60^0 = \sqrt{3}/2$ and $cos 60^0 = 1/2$.

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Without some additional information, you quite simply can't. The closest you can come is using the identity $$\cos^2\theta=\frac1{\tan^2\theta+1}$$ (which is derived from the Pythagorean identity, and holds wherever $\tan\theta$ is defined), from which you can determine that $$\cos\theta=\pm\sqrt{\frac1{1+\tan^2\theta}}.$$ Now, if you have an additional assumption, such as that $\theta$ is acute, then this becomes $$\cos\theta=\sqrt{\frac1{1+\tan^2\theta}},$$ at which point you can use this together with the angle sum identity for cosine to solve your problem

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