Question

sina+sinb=sqrt of 3/2,cosa+cosb=sqrt of 1/2 then 'a+b'?

Answer 1

use the identity :
cos(a+b) = cosacosb - sinasinb
let's see what happend if you subtitute all the values given

Answer 2

so far i got this sinasinb+cosacosb=0

Answer 3

that's not i mean
first, distribute all values given into formula above

Answer 4

opsss, i read the equation wrong. sorry, may bad

Answer 5

I have no idea how to approach this question, I thought of squaring the two equations and adding them together so I can cancel out some values

and I forgot to mention, value of a and b is larger than 0 and smaller than 180

Answer 6

yes, that's nice idea!
sina+sinb=sqrt of 3/2 -> sin^2 a + sin^2 + 2sinasinb = 3/2 or
=>1 + 2sinasinb = 3/2
=> 2sinasinb = 3/2 - 1
=> 2sinasinb = 1/2
=> sinasinb = 1/4

cosa+cosb=sqrt of 1/2 --> cos^2 a + cos^2 b + 2cosacosb = 1/2 or
=>1 + 2cosacosb = 1/2
=> 2cosacosb = 1/2 - 1
=> 2cosacosb = -1/2
=> cosacosb = -1/4

then use the identity like i said above :
cos(a+b) = cosacosb - sinasinb
therefore,
cos(a+b) = -1/4 - 1/4
cos(a+b) = -1/2
cos(a+b) = cos120
so, a+b = 120
with a,b < 180

Answer 7

i hope that make sense :)

Answer 8

hm I understand everything but Im not sure where this 1 came from.. 1 + 2sinasinb = 3/2

Answer 9

Hint: Use this identity then divide equations.
sinA + sinB = sin(A+B/2)cos(A-B/2)
cosA + cosB = cos(A+B/2)cos(A-B/2)

Answer 10

hmmm.. yes, that still wrong. i forgot to add both equation
correction ;
sin^2 a + sin^2 b + 2sinasinb + cos^2 a + cos^2 b + 2cosacosb = 3/2 + 1/2

(sin^2 a + cos^2 a) + (sin^2 b + cos^2 b) + 2(sinasinb + cosacosb) = 2
1+1 + 2cos(a-b) = 2
2 + 2cos(a-b) = 2
2cos(a-b) = 2 - 2
2cos(a-b) = 0
cos(a-b) = 0
cos(a-b) = cos90
a-b = 90

hmmm... where a+b ? --"

Answer 11

@SAValkyrie although squaring is fine sometimes, it should be avoided as far as possible as it'll lead to surplus solutions. You'll have to plug back the values and crosscheck.

Answer 12

so you would get tan(A+B/2)=sqrt(3)

Answer 13

YEP. And you know when tanC = sqrt3
It means its in the first quad,
And it is equal to 60 degrees.
So A+B = 120 degrees

Answer 14

Oh I see that answer makes sense, although I tried doing the squaring thingy because in the solution i got, it said to square both equations and add them together. But that was all that was written in the solution so I was wondering if there's a way to solve this without using different identities

Answer 15

I am actually in Grade 10 so I havent learned those identities in school yet, the question was actually from math competition.. ;;

Answer 16

Hm Thanks for all the help, I guess I would have to work on memorizing the identities of the trigonometry from now on!!

Answer 17

ah i got it now

we have a-b = 90 ------> a = b+90
take the 1st equ :
sina+sinb=sqrt of 3/2
subtitute a = b + 90
sin(b+90) + sinb = sqrt of 3/2
cosb + sinb = sqrt(3/2)

take the 2nd equ :
cosa+cosb=sqrt of 1/2
cos(b+90) + cosb = sqrt(1/2)
-sinb + cosb = sqrt(1/2)

divide both equation, we get
(cosb + sinb)/ (-sinb + cosb) = sqrt(3/2)/ sqrt(1/2)
(cosb + sinb)/ (cosb - sinb) = sqrt(3)
cosb + sinb = sqrt(3)(cosb - sinb)
cosb + sinb = sqrt(3)cosb -sqrt(3) sinb
(1+sqrt(3))sinb = (1+sqrt(3))cosb
sinb/cosb = 1
tanb = 1
b = 45

a-b = 90
a - 45 = 90
a = 45 + 90
a = 135

so, a+b = 135 + 45 = 180

Answer 18

\[\sin a+\sin b=\sqrt{\frac{ 3 }{ 2 }},\cos a+\cos b=\sqrt{\frac{ 1 }{ 2}}\]
squaring and adding,
\[\sin ^{2}a+\sin ^{2}b+2\sin a \sin b+\cos ^{2}a+\cos ^{2}b+2\cos a \cos b=\frac{ 3 }{2 }+\frac{ 1 }{2 }=2\]
\[\left( \sin ^{2}a+\cos ^{2}a \right)+\left(\sin ^{2}b+\cos ^{2}b \right)+2\left( \cos a \cos b+\sin a \sin b \right)=2\]
\[1+1+2\cos \left( a-b \right)=2\]
\[\cos \left( a-b \right)=0\]
Again
\[\left( \cos ^{2}a-\sin ^{2}a \right)+\left( \cos ^{2}b-\sin ^{2} b\right)+2\left( \cos a \cos b-\sin a \sin b \right)=\frac{ 1 }{2 }-\frac{ 3 }{ 2}=-1\]
\[\cos 2a+\cos 2b+2\cos \left( a+b \right)=-1\]
\[2\cos \frac{ 2a+2b }{ 2 }\cos \frac{ 2a-2b }{2 }+2\cos \left( a+b \right)=-1\]
\[2\cos \left( a+b \right)\cos \left( a-b \right)+2\cos \left( a+b \right)=-1\]
0+2cos(a+b)=-1
\[\cos \left( a+b \right)=\frac{ -1 }{2 }=-\cos \frac{ \pi }{ 3 }=\cos \left( \pi-\frac{ \pi }{ 3 } \right)=\cos \frac{ 2\pi }{ 3 }\]
\[a+b=\frac{ 2\pi }{ 3}\]

Answer 19

cosacosb+sinasinb how did this became cos(a-b)?? Is that one of the identities?

Answer 20

cos(a-b)=cosacosb+sinasinb

Answer 21

Oh I see, thanks for the answer, sadly I can only choose 1 best response D:

Answer 22

i gave for surjithayer, and you have to give to @shiddantsharan :)