Trig. Help? (1 Viewer)

[jayy100]

ok, i can't seem to work these questions out! There in the 3u year 11 Cambridge textbook and in the 'easy' section LOL

if q1. tan(a)=1/2 and tan(b)=1/3, find tan(a+b) (Ans.is 1)
q2. If cos(a)= 4/5 and sin(b)=12/13, where a and b are both acute, find sin(a+b) (Ans.is 63/65)

i think im over thinking these! :S any help would be much appreciated :]

 

[Bored Of Fail 2]

use double angle formulas / sum of angle formulas


q1. tan(a+b) = [ tan(a) + tan(b) ] / (1-tan(a)tan(b) )
q2. sin ( a+b) = sin(a)cos(b) + cos(a)sin(b)



if cos(a) = 4/5 , draw a general triangle and use a^2 =b^2 +c^2 , and you get sin(a) = 3/5

if sin(b) = 12/13, then cos(b) = 5/13

note we are told than a and b are acute ( in the first quadrant ) so all the ratios are positive

then sub in

 

[SpiralFlex]

Q1) tan (a + b) = (tan a + tan b)/(1 - tan a * tan b)

= (1/2 +1/3)/(1-(1/2 * 1/3)

= 1

Q2) Draw the triangle for Cos a = 4/5
Sin a should be 3/5 (by Pythag.)

Similarly get cosb from sinb. Therefore, cosb = 5/13

Sin (a + b) = sina*cosb + cosa*sinb

= 3/5 * 5/13 + 4/5 * 12/13

= 63/65

 

[jayy100]

OMG. THANKYOU LOL --"

errrr.. i was subbing in the sin^-1 functions into the calculator ><"

thanks guys. appreciate it heaps

 

[SpiralFlex]

No worries, Bored beat me to it.

Are we the only ones doing Cambridge at 10:30 pm?

 

[Bored Of Fail 2]

No worries, Bored beat me to it.

Are we the only ones doing Cambridge at 10:30 pm?

nah man, the others are waiting till 4am in the morning beofre they do cambridge