Question-161079

Question Number 161079 by 0731619 last updated on 11/Dec/21

Commented by mr W last updated on 11/Dec/21

m u s t > 0 a n d < 1, t h e r e f o r e o n l y \frac{1}{16}

c a n b e, i f o n e a n s w e r s h o u l d b e t r u e .

Commented by 0731619 last updated on 11/Dec/21

s o l u t i o n s i r ?

Commented by infinityaction last updated on 23/Apr/22

u s e f o r m u l a

\frac{1}{4} \cos 3 x = \cos (60 - x) \cos (x) \cos (60 + x)

p u t x = 25

\frac{1}{4} \cos 75 = \cos 35 \cos 25 \cos 85

p = \cos 15 (\frac{1}{4} \cos 75)

p = \frac{2}{8} \cos 15 \sin 15

p = \frac{1}{8} \times \sin 30

p = \frac{1}{16}

Answered by TheSupreme last updated on 11/Dec/21

\frac{1}{4} [c o s 40 + c o s (10)] [c o s (50) + c o s (120)] =

= \frac{1}{4} [c o s (40) c o s (50) + c o s (10) c o s (50) - \frac{1}{2} c o s (40) - \frac{1}{2} c o s (10)]

= \frac{1}{4} [\frac{1}{2} (c o s (90) + c o s (10)) + \frac{1}{2} (c o s (60) + c o s (40)) - \frac{1}{2} (c o s (10) + c o s (40)]

\frac{1}{4} [\frac{1}{2} c o s (60)] = \frac{1}{16}

Commented by mr W last updated on 12/Dec/21

g o o d s i r! t h a n k s!

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