tan3a-tan2a-tana = tan3a tan2a tana

It's a bit of work, but...

Write out each tangent as the sine/cosine equivalent, and then use trig identities to convert the terms that are multiples of 'a' into sina and cosa equivalents. For example: sin2a = 2sinacosa and cos2a = cos^2a-sin^2a. I will leave it to you to figure out how to converyt sin3a and cos3a into terms of sina and cosa. Once you get it into a form that has only sina and cosa terms, you'll see it works out.

Quote:

---

Originally Posted by ebaines

It's a bit of work, but...

Write out each tangent as the sine/cosine equivalent, and then use trig identities to convert the terms that are multiples of 'a' into sina and cosa equivalents. For example: sin2a = 2sinacosa and cos2a = cos^2a-sin^2a. I will leave it to you to figure out how to converyt sin3a and cos3a into terms of sina and cosa. Once you get it into a form that has only sina and cosa terms, you'll see it works out.

---

I an still confused... I tried doing it the way you suggested but don't get the answer... do you think this way actually works.

Quote:

---

Originally Posted by zaibjaffer

I an still confused .....I tried doing it the way you suggested but dont get the answer......do you think this way actually works.

---

Yes, it does.

I started by combining the left hand side over a common denominator, which turns out to have the same denominator as the right hand side. So then it's merely a job of showing that both numerators are the same. In other words:

sin(3a)*cos(2a)*cos(a) - sin(2a)*cos(3a)*cos(a) - sin(a)*cos(3a)*cos(2a) = sin(3a)*sin(2a)*sin(a)

At this point, you substitute for sin(3a), cos(3a), sin(2a) and cos(2a) with identities that are purely in sin(a) and cos(a). Do you know what those identities are?

Quote:

---

Originally Posted by ebaines

Yes, it does.

I started by combining the left hand side over a common denominator, which turns out to have the same denominator as the right hand side. So then it's merely a job of showing that both numerators are the same. In other words:

sin(3a)*cos(2a)*cos(a) - sin(2a)*cos(3a)*cos(a) - sin(a)*cos(3a)*cos(2a) = sin(3a)*sin(2a)*sin(a)

At this point, you substitute for sin(3a), cos(3a), sin(2a) and cos(2a) with identities that are purely in sin(a) and cos(a). Do you know what those identities are?

---

Thanks for helping me it worked...
I have an other question now
2sin5xcos4x-sinx=sin9x

You shouldn't post a new question in the same thread as an old. Please start a new question.

Ha ha. Very funny:-)