Question

Finding (i)`tanAtanB`, (ii)`tan(A+B)`, (iii)`sin((A+B)/2)` using Addition Formulae?

Given `cos(A+B) = 3/5` and `cos A cos B = 7/10`, where A and B are both acute angles, find the value of
(i)`tanAtanB`,
(ii)`tan(A+B)`,
(iii)`sin((A+B)/2)`.

Answers:
(i) `1/7`
(ii) `3/4`
(iii) `sqrt(5)/5`

Answer 1

Those are right except (ii) is inverted. `tan(A+B)` should be `4/3` as `sin(A+B)=4/5` and `cos(A+B)= 3/5`.

Explanation:

Fun. Given `cos (A+B) = 3/5 quad and quad cos A cos B=7/10`

Let's review the relevant identities.

`cos(A+B)=cos A cos B - sin A sin B`

`sin A sin B = cos A cos B -cos(A+B) = 7/10 - 3/5 = 1/10`

`tanA tan B = {sin A sin B}/{cos A cos B} = {1/10}/{ 7/10 } = 1/7 quad` choice (i)

`cos ^2(A+B) + sin^2(A+B) = 1`

`sin(A+B)=\pm \sqrt{1-(3/5)^2} = \pm 4/5`

`A` and `B` are acute, `A+B<180^circ` so a positive sine:

`sin(A+B)=4/5`

`tan(A+B) = sin(A+B)/ cos(A+B) = {4/5}/{3/5} = 4/3 quad` NONE OF THE ABOVE

One double angle formula is `cos(2x)=1-2 sin ^2 x` so

`sin( ( A+B)/2 ) = \pm \sqrt{1/2 (1 - cos(A+B))}`

The average of `A` and `B` is acute, so we choose the positive sign.

`sin( ( A+B)/2 ) = + \sqrt{1/2 (1 - 3/5)) = 1/\sqrt{5} quad` choice (iii)

One of three wrong, B-.

Answer 2

Kindly refer to the Explanation Section.

Explanation:

Given that `cos(A+B)=3/5`.

`:. cosAcosB-sinAsinB=3/5`.

`:. 7/10-sinAsinB=3/5`.

`:. sinAsinB=7/10-3/5=1/10`.

`:. (sinAsinB)/(cosAcosB)=(1/10)/(7/10)`.

Hence, `tanAtanB=1/7.............."[Ans."(i)]`.

Given that, `0 lt A lt pi/2, 0 lt B lt pi/2`.

Adding, `0 lt (A+B) lt pi`.

`:. (A+B) in Q_1uuQ_2`.

But, `cos(A+B)=3/5 gt 0`.

`:. (A+B) in Q_1`.

Now, `sin^2(A+B)=1-cos^2(A+B)=1-(3/5)^2=16/25`.

`:. sin(A+B)=+-4/5;" but, because, "(A+B) in Q_1,`

`sin(A+B)=+4/5`.

`:.tan(A+B)=sin(A+B)/cos(A+B)=(4/5)/(3/5)=4/3..."[Ans."(ii)]`.

Finally, to find `sin((A+B)/2)," let, "(A+B)/2=theta.`

`:. cos(A+B)=cos2theta=3/5`.

`"Now, "cos2theta=3/5 rArr cos(theta+theta)=3/5`.

`:. costhetacostheta-sinthetasintheta=3/5...[because,"Addition Formula]"`

`:. cos^2theta-sin^2theta=3/5, i.e.,`

`(1-sin^2theta)-sin^2theta=3/5, or,`

`1-2sin^2theta=3/5 rArr sin^2theta=1/2(1-3/5)=1/5`.

`:. sintheta=+-1/sqrt5`

Since, `(A+B)=2theta` lies in `Q_1," so does "theta=(A+B)/2`.

`:. sintheta=sin((A+B)/2)=+1/sqrt5=+sqrt5/5......"[Ans."(iii)]`.