Question

prove that: cos28-cos58-sin2=0​

Answer 1

Answer:

1+cos56+cos58−cos66=mcos20cos29sin33

=mcos28cos29sin33

=

2

m

[2cos28cos29]sin33

=

2

m

[cos57+cos1]sin33

=

4

m

[2cos57sin33+2cos1sin33]

=

4

m

[sin(90)+(sin24)+sin34−sin32]

=

4

m

[1−sin24

o

+sin34

o

−sin32

o

]

=

4

m

[1−cos66

o

cos56

o

−cos58

o

]

⇒

4

m

=1⇒m=4

Hence the value of m is 4