A maths - Trigonometric functions

bonkysleuth SG

A is an obtuse angle and sin A = sq.rt 3/2. Find the value of the following

cos 1/2A

Here's my working:

cos^2 (0.5A) - sin^2(0.5A)

=2 cos^2 (0.5A) -1

cos A = 2 cos^2 (0.5A) -1

-0.5 = 2 cos^2 (0.5A) -1

0.25 = cos^2 (0.5A)

cos 0.5A = sqrt. 1/4

= 0.5 or -0.5(rej)

i do not understand why we must reject the negative value. it is an obtuse angle,so should the positive one be rejected instead since cos is negative in 2nd quad?

eagle SG

A is an obtuse angle, means A is between 90 and 180 degrees.

Means 0.5A is between 45 to 90 degrees

In this region, cosine is positive.

Hence you reject the negative value.

bonkysleuth SG

A is an obtuse angle, means A is between 90 and 180 degrees.

Means 0.5A is between 45 to 90 degrees

In this region, cosine is positive.

Hence you reject the negative value.

Thanks. Here is one question which I can't do. I would be glad if you were to assist.

prove the identity

1 + tan^2 A / 1 -tan^2 A = sec 2A

May I ask whether the original 'A' is A or 2A? cuz if it's A, then we will use 0.5A when applying the cos 2A method and we use A if the original one is 2A (HALF of the original term)

thanks again.

Mikethm SG

LHS

= (1+sin^2 A/cos^ 2A)/(1-sin^2 A/cos^ 2 A)

=[(cos^2 A+sin^2 A)/cos^ 2 A]/[(cos^2 A-sin^2 A)/cos^ 2 A]

= (1/cos^2 A)(cos^2 A/cos 2A)

= 1/cos 2A

= sec 2A

=RHS