sin[sin3+sin5]=cos[cos3-cos5] Establish the identity I used the sum to product formula so far
is that a literal 3 and 5? or are those exponents?
that is \(\bf \large sin[sin(3)+sin(5)]=cos[cos(3)-cos(5)] ?\)
yes it's just three and five, they have thetas after them as well @jdoe0001
hmmm
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJzaW4oc2luKDN4KStzaW4oNXgpKSIsImNvbG9yIjoiI0QxMUYxRiJ9LHsidHlwZSI6MCwiZXEiOiJjb3MoY29zKDN4KS1jb3MoNXgpKSIsImNvbG9yIjoiIzE4MzlERSJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi02LjUxOTk5OTk5OTk5OTk5OSIsIjYuNDgiLCItMy45OCIsIjQuMDIiXX1d hmmmm.....so... as you can see. hmmm I tried a few dead ends ...so....notice the graph
I have to show how the left side equals the right without changing the right hand side
Using trig functions
well... if you notice the graph you'd end up in dead ends though
Is this what you are trying to show?\[\sin(\theta)\left(\sin(3\theta)+\sin(5\theta)\right)=\cos(\theta)\left(\cos(3\theta)-\cos(5\theta)\right)\]
Yes
OK - notice how in this identity we have \(\theta,3\theta,5\theta\) where we notice that:\[\theta+3\theta=4\theta\]and:\[5\theta-\theta=4\theta\]
Can you see how to make use of this fact?
Think about the cosine addition and subtraction formulas
I tried using the sum to product formula
From what I showed you above we can write:\[\theta+3\theta=5\theta-\theta\]Now just take cosines of both sides to get:\[\cos(\theta+3\theta)=\cos(5\theta-\theta)\] Finally expand each of these using the cosine addition/subtraction formulas, rearrange and you will get to the desired identity.
I dont understand how you got that in the previous part @asnaseer
1+3=5-1=4
Do you understand?
Isnt the sin on the outside multiplied
Yes it is, and if you expand this out as I suggested then you will get to your identity:\[\cos(\theta+3\theta)=\cos(5\theta-\theta)\]
Have you been asked to specifically use the sum to product formulas?
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