IfA=340°, prove that
2cosA/2 = -(v1+sinA) - (v1-sinA)
Steve, drwls and I have answered quite a few of your questions based on the same type of trig expansions in the last few weeks.
You have not given a single sign showing any effort on your part to solve these problems.
Show us how far you got in solving these problems and perhaps we can give you a hint on how to continue.
There is also no indication that you have even looked at our solutions to your problems.
Good comment, Reiny. I generally eschew working on homework dumps, but I have always enjoyed the many ways trig functions can be combined. Some of those tan-sec identities in the previous batch stumped me, though. Today's problems were pretty straightforward.
Still, Kewal, it would be nice to see that after all this time you have some idea where to start. Show us some effort, even if it's totally wrong, and we'll set you straight. Unless I happen to have some quiet time, I probably won't be tackling any more of these without some sign of participation from you.
To prove the equation 2cos(A/2) = -(v1 + sinA) - (v1 - sinA) when A = 340°, we need to break it down step by step and simplify both sides of the equation.
First, let's simplify the left side of the equation, which is 2cos(A/2):
1. Start with the double angle formula for cosine: cos(2θ) = 2cos^2(θ) - 1.
2. Substitute A/2 for θ in the double angle formula: cos(A) = 2cos^2(A/2) - 1.
3. Divide both sides of the equation by 2: (1/2)cos(A) = cos^2(A/2) - 1/2.
4. Rearrange the equation to isolate cos(A/2): cos(A/2) = √[(1/2)cos(A) + 1/2].
Now, let's simplify the right side of the equation, which is -(v1 + sinA) - (v1 - sinA). We'll combine like terms as we go:
1. Distribute the negative sign: -(v1 + sinA) - (v1 - sinA) = -v1 - sinA - v1 + sinA.
2. Combine like terms: -v1 - v1 = -2v1.
3. Simplify further: -2v1.
Now, we can compare both sides of the equation:
2cos(A/2) = -(v1 + sinA) - (v1 - sinA)
√[(1/2)cos(A) + 1/2] = -2v1
Since the equation involves functions and variables, we can't directly solve for v1 using algebraic methods. However, we have successfully simplified both sides of the equation.