## anonymous one year ago anyone pls help me .....I will become ur fan and give u a medal?????

1. anonymous

cos (a+45)cos (a-45)=1/2 -sin^2 a

2. anonymous

that is the real q

3. zepdrix

Hey Panda :) Is this an identity that we want to prove? Or are we simply trying to solve for a?

4. anonymous

ithis is an identity

5. zepdrix

$\large\rm \cos(a+45^o)\cos(a-45^o)=\frac{1}{2}-\sin^2a$Hmm there is probably some shortcut for doing this.. I can't seem to remember though. I guess what I would do is apply your Angle Sum Formula to both cosines.$\large\rm \cos(\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$$\large\rm \cos(\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$

6. anonymous

then????

7. zepdrix

So on the left side, applying both identities would give you$\large\rm (\cos a \cos45-\sin a \sin45)(\cos a \cos45+\sin a \sin45)$Recognize that here we have conjugates being multiplied together. So we'll get the difference of squares,$\large\rm \cos^2a \cos^245-\sin^2a \sin^245=\frac{1}{2}-\sin^2a$

8. zepdrix

Recall that at 45 degrees, you're getting the same value for sine and cosine, sqrt2/2.$\large\rm \left(\frac{\sqrt2}{2}\right)^2\cos^2a -\left(\frac{\sqrt2}{2}\right)^2\sin^2a=\frac{1}{2}-\sin^2a$Which simplifies a bit,$\large\rm \frac{1}{2}\cos^2a -\frac{1}{2}\sin^2a=\frac{1}{2}-\sin^2a$Which is quite a bit closer to where we want to be :)

9. zepdrix

What do you think? :) Too confusing doing it this way?

10. anonymous

yup

11. anonymous

another q pls...how to solve these q

12. LynFran

ok we need to express in terms of $R \cos(x+\alpha)$to find R$\sqrt{2+1}$so R=$\sqrt{3}$and to get alpha we find tan of it so $\tan^{-1}\frac{ 1 }{ 2 }=26.565$now we can express it in the given form... $R \cos(x+\alpha)$$\sqrt{3}\cos(x+26.565)$and then to solve the second part we let $\sqrt{3}\cos(x+26.565)=1$can u continue from there?

13. anonymous

HOW DID U GET THE R

14. LynFran

i took the amplitude of 2cosx..which is 2...and the amplitude of sinx..which is 1...and then $\sqrt{2+1}$

15. anonymous

OOOH

16. anonymous

I CANNOT SOLVE IT........

17. LynFran

$\sqrt{3}\cos (x+26.565)=1$$\cos(x+26.565)=\frac{ 1 }{ \sqrt{3} }$we rationalise the denominator $\cos(x+26.565)=\frac{ \sqrt{3} }{ 3 }$we take cos inverse..$(x+26.565)=\cos^{-1} \frac{ \sqrt{3} }{ 3 }+2n \pi$ok since ur using degree and not radians we have$(x+26.565)=\pm54.736+360$$x=-26.565\pm54.736+360$..so can u find the solution from here? now u must look at the given $0\le x \le 180$

18. LynFran

can u solve for x from there?

19. LynFran

ok so $x=-26.565+54.736$now the +360 is the period here thats y we continue to add the period for given domain cause ur answer would vary of that domain so.. $x=28.171$now if we add the period which is 360 here we get$x=28.171+360$$x=388.171$now this 388.171 is out of the given domain so thats not a solution so we have 1 solution to this problem for the given domain $x=28.171$...now if we check the other $x=-26.565-54.736$$x=-81.301$...this is out of domain if we add 360 we get $x=-81.301+360$$x=278.699$so it would not be a solution... the only solution is $x=28.171$

20. anonymous

SORRY FOR THE LATE REPLY....I JUST ATE MY BREAKFAST...

21. anonymous

OOOH NOW I GET IT ..... TQ VERY MUCH....

22. LynFran

ok