So much sine
The numerical value of ( sin 2 0 ∘ ) ( sin 4 0 ∘ ) ( sin 6 0 ∘ ) ( sin 8 0 ∘ )
Can be expressed in the form b a . Find a + b .
The answer is 19.
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2 solutions
the upper formula? where did you get that? i never heard.
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Oh!, you should know these formulas. These are very useful. Here is a link for this formula. Its good that you have learnt these now. There are some similar formulas too, like 4 c o s ( x ) c o s ( 6 0 − x ) c o s ( 6 0 + x ) = c o s ( 3 x ) and t a n ( x ) t a n ( 6 0 − x ) t a n ( 6 0 + x ) = t a n ( 3 x ) Stt that the number 4 appears only in s i n and c o s ′ s formulas. When you divide then, you get a wonderful result of t a n ′ s formula..
LaTeX tip: use \sin and \cos for better looking sines and cosines. sin x cos x .
But wat about sin(20) .. where did you solve it
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We have s i n ( x ) s i n ( 6 0 − x ) s i n ( 6 0 + x ) = s i n ( 3 x ) 4 1 So, in this case when x = 2 0 , we have ,,,,
s i n ( 2 0 ) s i n ( 6 0 − 2 0 ) s i n ( 6 0 + 2 0 ) = 4 1 s i n ( 3 × 2 0 ) ..
I Hope, its clear now
As:
sin(a)sin(b)=1/2[cos(a-b)-cos(a+b)]
cos(a)cos(b)=1/2[cos(a-b)+cos(a+b)].
We can do:
sin20sin40sin60sin80=1/4 *(cos20-cos60)(cos20-cos140).
As cos140=-cos40 and cos100=-cos80:
1/4 (cos²20-cos20cos60+cos20cos40-cos40cos60)
=1/8 (1+cos40-cos40+cos80+cos20+cos60-cos20-cos100)
=1/8 (1+1/2)=3/16
Then:
3/16=a/b
a+b=19.
We know that s i n ( x ) s i n ( 6 0 − x ) s i n ( 6 0 + x ) = 4 1 s i n ( 3 x ) Using this property, we get s i n ( 6 0 ) s i n ( 2 0 ) s i n ( 6 0 − 2 0 ) s i n ( 6 0 + 2 0 ) = 4 1 s i n 2 ( 6 0 )
Now, this is equal to
= 1 6 3 = b a Therefore (a+b=19