1. ## [SOLVED]Trigonometric identities.

Could any kind souls help me with these questions? I am stuck!

1)Prove the identity: cos(a+b) cos(a-b) = cos^2a -sin^2b
for this, i am stuck at LHS = cos^2a cos^2b - sin^2a sin^2b .

2)Prove the identity tanA tan(A-60) + tanA tan(A+60) + tan(A-60) tan(A+60) = -3 ,
For this,im stuck with LHS = -3tan^2A . Please correct me if this is already wrong cause i assumed A to be 180 degrees.

3)If A and B are the principal values of arccos(3/sqrt10) and arcsin(1/sqrt5) respectively,show that A+B = pi/4 (I have no idea how to start this,i've never seen this question before!)

4) Prove that if A+B+C = 180 degrees(sum of angles in a triangle),
cotAcotB + cotBcotC + cotCcotA =1 holds true.

(for this, i've tried to use 1/tanAtanB... and also converting it to cosA/sinA... but both of these methods got me into infinite loops and some nice symmetry too. Haha!)

All help is greatly appreciated!

2. Originally Posted by arccos
Could any kind souls help me with these questions? I am stuck!

[snip]
3)If A and B are the principal values of arccos(3/sqrt10) and arcsin(1/sqrt5) respectively,show that A+B = pi/4 (I have no idea how to start this,i've never seen this question before!)

[snip]
Consider $\displaystyle \cos (A + B) = \cos (A) \cos(B) - \sin(A) \sin(B)$ where $\displaystyle \cos(A) = \frac{3}{\sqrt{10}} \Rightarrow \sin(A) = ....$ and $\displaystyle \sin(B) = \frac{1}{\sqrt{5}} \Rightarrow \cos(B) = ....$.

Once you substitute these values you should get $\displaystyle \cos(A + B) = \frac{1}{\sqrt{2}} \Rightarrow A + B = \frac{\pi}{4}$.

3. Originally Posted by arccos
1)Prove the identity: cos(a+b) cos(a-b) = cos^2a -sin^2b
for this, i am stuck at LHS = cos^2a cos^2b - sin^2a sin^2b .
Good start. Now change $\displaystyle \cos^2{b}$ with $\displaystyle 1-\sin^2{b}$.

4. Thanks for the inputs! I have solved qn 1 and 3 with your help. Now i'm left with 2 & 4.

5. Originally Posted by arccos
Thanks for the inputs! I have solved qn 1 and 3 with your help. Now i'm left with 2 & 4.
Q2:

Use the following identitites:

$\displaystyle \tan(a+b) = \dfrac{\tan(a)+ \tan(b)}{1- \tan(a) \; \tan(b)}$

$\displaystyle \tan(a-b) = \dfrac{\tan(a)-\tan(b)}{1+\tan(a) \; \tan(b)}$

$\displaystyle tan(60) = tan(\frac{\pi}{3})=\sqrt{3}$

and then keep simplifying to get -3.

6. Originally Posted by harish21
Q2:

Use the following identitites:

$\displaystyle \tan(a+b) = \dfrac{\tan(a)+ \tan(b)}{1- \tan(a) \; \tan(b)}$

$\displaystyle \tan(a-b) = \dfrac{\tan(a)-\tan(b)}{1+\tan(a) \; \tan(b)}$

$\displaystyle tan(60) = tan(\frac{\pi}{3})=\sqrt{3}$

and then keep simplifying to get -3.

How can i use the addition formulaes for tangent in the question? i cant see any relation. would you please enlighten me?

7. tanA tan(A-60) + tanA tan(A+60) + tan(A-60) tan(A+60)

$\displaystyle =\bigg(\tan(A) \times \dfrac{\tan(A)-\tan(60)}{1+\tan(A)\;\tan(60)}\bigg) \;+ \; \bigg(\tan(A) \times \dfrac{\tan(A)+\tan(60)}{1-\tan(A)\;\tan(60)}\bigg)+ \bigg(\dfrac{\tan(A)-\tan(60)}{1+\tan(A)\;\tan(60)} \times \dfrac{\tan(A)+\tan(60)}{1-\tan(A)\;\tan(60)}\bigg)$

$\displaystyle =\bigg(\tan(A) \times \dfrac{\tan(A)-\sqrt{3}}{1+\sqrt{3}\tan(A)}\bigg) \;+ \; \bigg(\tan(A) \times \dfrac{\tan(A)+\sqrt{3}}{1-\sqrt{3}\tan(A)}\bigg)+ \bigg(\dfrac{\tan(A)-\sqrt{3}}{1+\sqrt{3}\tan(A)} \times \dfrac{\tan(A)+ \sqrt{3}}{1-\sqrt{3}\tan(A)}\bigg)$

$\displaystyle =\bigg(\dfrac{\tan^{2}(A)-\sqrt{3} \tan(A)}{1+\sqrt{3}\tan(A)}\bigg) + \bigg( \dfrac{\tan^{2}(A)+\sqrt{3}\tan(A)}{1-\sqrt{3}\tan(A)}\bigg)+\bigg(\dfrac{\tan^{2}(A)-3}{(1+\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))}\bigg)$

now finish it..

8. Originally Posted by harish21
tanA tan(A-60) + tanA tan(A+60) + tan(A-60) tan(A+60)

$\displaystyle =\bigg(\tan(A) \times \dfrac{\tan(A)-\tan(60)}{1+\tan(A)\;\tan(60)}\bigg) \;+ \; \bigg(\tan(A) \times \dfrac{\tan(A)+\tan(60)}{1-\tan(A)\;\tan(60)}\bigg)+ \bigg(\dfrac{\tan(A)-\tan(60)}{1+\tan(A)\;\tan(60)} \times \dfrac{\tan(A)+\tan(60)}{1-\tan(A)\;\tan(60)}\bigg)$

$\displaystyle =\bigg(\tan(A) \times \dfrac{\tan(A)-\sqrt{3}}{1+\sqrt{3}\tan(A)}\bigg) \;+ \; \bigg(\tan(A) \times \dfrac{\tan(A)+\sqrt{3}}{1-\sqrt{3}\tan(A)}\bigg)+ \bigg(\dfrac{\tan(A)-\sqrt{3}}{1+\sqrt{3}\tan(A)} \times \dfrac{\tan(A)+ \sqrt{3}}{1-\sqrt{3}\tan(A)}\bigg)$

$\displaystyle =\bigg(\dfrac{\tan^{2}(A)-\sqrt{3} \tan(A)}{1+\sqrt{3}\tan(A)}\bigg) + \bigg( \dfrac{\tan^{2}(A)+\sqrt{3}\tan(A)}{1-\sqrt{3}\tan(A)}\bigg)+\bigg(\dfrac{\tan^{2}(A)-3}{(1+\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))}\bigg)$

now finish it..

I simplified it to 8tan^2A - 3 / 1 -3tan^2A and this was as much as I could go. I got to here by multiplying the first 2 brackets of fractions together and i found that it had the same denominator as the last bracket.Expanded(really long) and simplified to that. Am i doing it wrong?

9. Originally Posted by arccos
I simplified it to 8tan^2A - 3 / 1 -3tan^2A and this was as much as I could go. I got to here by multiplying the first 2 brackets of fractions together and i found that it had the same denominator as the last bracket.Expanded(really long) and simplified to that. Am i doing it wrong?
Check your calculation again..you should be getting:

$\displaystyle \dfrac{9\tan^{2}(A)-3}{1-3\tan^{2}(A)}$

10. Originally Posted by harish21
Check your calculation again..you should be getting:

$\displaystyle \dfrac{9\tan^{2}(A)-3}{1-3\tan^{2}(A)}$

Testing latex:$\displaystyle (tan\^{}2A-sqrt3tanA)(1-sqrt3tanA)+ (tan\^{}2A + sqrt3tanA)(1+sqrt3tanA) =tan\^{}2A -sqrt3tan\^{}3A -sqrt3tanA +3tan\^{}2A + tan\^{}2A +sqrt3tan\^{}3A + sqrt3tanA +3tan\^{}2A$

Normal text:
(tan^2A-sqrt3tanA)(1-sqrt3tanA)+ (tan^2A + sqrt3tanA)(1+sqrt3tanA)
=tan^2A -sqrt3tan^3A -sqrt3tanA +3tan^2A + tan^2A +sqrt3tan^3A + sqrt3tanA +3tan^2A (just the numerator)

Its late here. I still see 8 only. :O

11. For the last one:

$\displaystyle A + B + C = 180$

$\displaystyle A + B = 180 - C$

$\displaystyle \tan(A+B) = \tan(180 - C)$

$\displaystyle \dfrac{\tan A + \tan B}{1 - \tan A\tan B} = \dfrac{\tan 180 - \tan C}{1 + \tan 180 \tan C}$

but tan 180 = 0;

$\displaystyle \dfrac{\tan A + \tan B}{1 - \tan A\tan B} = - \tan C$

$\displaystyle \tan A + \tan B = - \tan C + \tan A\tan B\tan C$

$\displaystyle \tan A + \tan B + \tan C = \tan A\tan B\tan C$

$\displaystyle \dfrac{1}{\tan A + \tan B + \tan C} = \dfrac{1}{\tan A\tan B\tan C}$

$\displaystyle \dfrac{1}{\tan A + \tan B + \tan C} = \cot A\cot B\cot C}$

now cross multiply, keeping in mind that:

$\displaystyle \cot A = \dfrac{1}{\tan A}$

12. Unknown008,thanks so much! I totally overlooked the fact that tan180(i wrote it as tan(pi) radians) = 0. Looks like I haven't got the special values at my fingertips yet.

13. You should have:

$\displaystyle \dfrac{(\tan^{2}(A)-\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))+(\tan^{2}(A)+\sqrt{3}\tan(A))(1+\ sqrt{3}\tan(A))+\tan^{2}(A)-3}{(1-\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))}$

the numerator would be:

$\displaystyle \tan^{2}(A)-\sqrt{3}\tan^{3}(A)-\sqrt{3}\tan(A)+3\tan^{2}(A)+\tan^{2}(A)+\sqrt{3}\ tan^{3}(A)+\sqrt{3}\tan(A)+3tan^{2}(A)+tan^{2}(A)-3$

$\displaystyle =\tan^{2}(A)+3\tan^{2}(A)+\tan^{2}(A)+3\tan^{2}(A) +\tan^{2}(A)-3$

$\displaystyle =9tan^{2}(A)-3$

14. Originally Posted by arccos
Unknown008,thanks so much! I totally overlooked the fact that tan180(i wrote it as tan(pi) radians) = 0. Looks like I haven't got the special values at my fingertips yet.
Well, I had a similar proof some time ago which stopped at the tanAtanBtanC = tanA + tanB + tanC, so, it was easier to get this right

And it's always good to check the special values now and again.

15. Originally Posted by harish21
You should have:

$\displaystyle \dfrac{(\tan^{2}(A)-\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))+(\tan^{2}(A)+\sqrt{3}\tan(A))(1+\ sqrt{3}\tan(A))+\tan^{2}(A)-3}{(1-\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))}$

the numerator would be:

$\displaystyle \tan^{2}(A)-\sqrt{3}\tan^{3}(A)-\sqrt{3}\tan(A)+3\tan^{2}(A)+\tan^{2}(A)+\sqrt{3}\ tan^{3}(A)+\sqrt{3}\tan(A)+3tan^{2}(A)+tan^{2}(A)-3$

$\displaystyle =\tan^{2}(A)+3\tan^{2}(A)+\tan^{2}(A)+3\tan^{2}(A) +\tan^{2}(A)-3$

$\displaystyle =9tan^{2}(A)-3$
Opps! I forgot about the last bracket.
Can you hint me on what to do with the simplified fraction?

EDIT: Oh i got it! I factored out the 3 for the numerator and multipled the fraction by -1 and the 3tan^2A -1 cancels out. Thanks so much!!!

### tanAtan(A 60°) tanAtan(A-60°) tan(A 60°)(tanA-60°) 3

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