D dabatman Member Joined Aug 13, 2016 Messages 122 Gender Undisclosed HSC N/A May 22, 2017 #1 Hi, Can anyone please find the exact value of this question please? Attachments Screen Shot 2017-05-22 at 2.48.00 pm.png 12.6 KB Views: 36
Squar3root realest nigga Joined Jun 10, 2012 Messages 4,927 Location ya mum gay Gender Male HSC 2025 Uni Grad 2024 May 22, 2017 #2 lhs = sin(x+30) + cos(x+30) now cos(x) = sin (90-x) so we get sin(x+30) + sin(90 - (x+30)) = sin(x+30) + sin (60-x) using sum to product we get 2sin( [x+30 +60+x ]/2) * cos ([x + 30 - (60 -x)]/2 = 2 sin(90/2) cos([2x-30]/2) = sqrt(2)cos(x-15)
lhs = sin(x+30) + cos(x+30) now cos(x) = sin (90-x) so we get sin(x+30) + sin(90 - (x+30)) = sin(x+30) + sin (60-x) using sum to product we get 2sin( [x+30 +60+x ]/2) * cos ([x + 30 - (60 -x)]/2 = 2 sin(90/2) cos([2x-30]/2) = sqrt(2)cos(x-15)
Squar3root realest nigga Joined Jun 10, 2012 Messages 4,927 Location ya mum gay Gender Male HSC 2025 Uni Grad 2024 May 22, 2017 #3 using the identity using product to sums first time using latex in long time no h8 pls