Derive this identity from the sum and difference formulas for cosine:
sinasinb=1/2[cos(a-b)cos(a+b)]
Start with the right-hand side since it is more complex.
Calculations:
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Reasons:
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prnt.sc/jnp0iz
PLEASE HELP ME!!!!!
2 answers
we know:
cos(a-b) = cosa cosb + sina sinb , and
cos(a+b) = cosa cosb - sina sinb
----------------------------------------- , let's subtract them
cos(a-b) - cos(a+b) = 2sina sinb
(1/2)( cos(a-b) + cos(a+b) = sina sinb , as required
cos(a-b) = cosa cosb + sina sinb , and
cos(a+b) = cosa cosb - sina sinb
----------------------------------------- , let's subtract them
cos(a-b) - cos(a+b) = 2sina sinb
(1/2)( cos(a-b) + cos(a+b) = sina sinb , as required
1 answer