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#

2 Answers
Dean R.
May 2, 2018

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-.

Ratnaker Mehta
May 2, 2018

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)]#.

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