# Trigonometric identities.

• Nov 4th 2010, 08:50 PM
arccos
[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(Rofl). Haha!)

All help is greatly appreciated!
• Nov 4th 2010, 09:35 PM
mr fantastic
Quote:

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 $\displaystyle \cos(A + B) = \frac{1}{\sqrt{2}} \Rightarrow A + B = \frac{\pi}{4}$.
• Nov 4th 2010, 09:48 PM
TheCoffeeMachine
Quote:

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 $\displaystyle 1-\sin^2{b}$.
• Nov 5th 2010, 01:13 AM
arccos
Thanks for the inputs! I have solved qn 1 and 3 with your help. Now i'm left with 2 & 4. (Rock)
• Nov 5th 2010, 06:17 AM
harish21
Quote:

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. (Rock)

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.
• Nov 5th 2010, 07:51 AM
arccos
Quote:

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?
• Nov 5th 2010, 08:07 AM
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..
• Nov 5th 2010, 08:19 AM
arccos
Quote:

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?
• Nov 5th 2010, 08:23 AM
harish21
Quote:

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:

$\displaystyle \dfrac{9\tan^{2}(A)-3}{1-3\tan^{2}(A)}$
• Nov 5th 2010, 08:32 AM
arccos
Quote:

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
• Nov 5th 2010, 08:32 AM
Unknown008
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:

$\displaystyle \cot A = \dfrac{1}{\tan A}$
• Nov 5th 2010, 08:37 AM
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.
• Nov 5th 2010, 08:41 AM
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:

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

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

$\displaystyle =9tan^{2}(A)-3$
• Nov 5th 2010, 08:43 AM
Unknown008
Quote:

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 (Smile)

And it's always good to check the special values now and again.
• Nov 5th 2010, 08:43 AM
arccos
Quote:

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!!!