Verifying an equation
\[\sin(x+\frac{ \pi }{ })−\cos(x+\frac{ \pi }{ 3 })=\sqrt{3}sinx\]
Can you please help me. I don't know my next step
oh crapp
its pi/6
\[\[\sin(x+\frac{ \pi }{ 6 })−\cos(x+\frac{ \pi }{ 3 })=sqroot(3) sinx\]\]
If you want to solve for x, expand the left hand side using the trig identity: sin(A+B) = sin(A)cos(B) + cos(A)sin(B) cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
okok let me do that
ok
so now i find all the exact values that I can get?
Well they are not asking you to solve the equation but just to verify. If you expand the left hand side and simplify you will notice it becomes sqrt(3)sin(x).
to simplify would i need to find some of the exact values ? because theres nothing i can do after expanding the LHS
well i don't see anything*
You can substitute the values for sin(pi/6), cos(pi/6), sin(pi/3), cos(pi/3) from the unit circle or trig tables.
ahhh ok that's what I meant. let me do that :p
im left with \[sinx \frac{ \sqrt{3} }{ 3 } - sinx \frac{ \sqrt{3} }{ 2 }\]
It should be \[\frac{ \sqrt(3) }{ 2 }\sin(x) + \frac{ \sqrt(3) }{ 2 }\sin(x) = \sqrt(3)\sin(x)\]
oh whoopesies I replaced the 2 with a 3
wait so sqroot3/2, where does the two go in the equation??? The denominator
\[\frac{ \sqrt(3) }{ 2 }\sin(x) + \frac{ \sqrt(3) }{ 2 }\sin(x) = 2 * \frac{ \sqrt(3) }{ 2 }sin(x) = \sqrt(3)\sin(x)\]
i don't understand how they merge
Not squared. Here we are ADDING two fractions. Not multiplying. If we are multiplying two identical terms we can square it. But when we are adding two identical terms we need to multiply by 2 or double it (not square it).
ahhh
5 + 5 = 2 * 5 = 10 a + a = 2 * a = 2a 1/4 + 1/4 = 2 * 1/4 = 1/2
since we are adding fraction with fraction
wow xD that was dumb of me. it was just because everything is like in big numbers
So we get sqrt(3) * sin(x) which is the right hand side. So we have verified what was asked.
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