Solve for theta
Tan5∅Tan3∅=1
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1) tan5x = sin5x/cos5x
2) tan3x = sin3x/cos3x
Add 1) and 2) and remember
A/B + C/D = (AD+BC)/(BD)
Using this relation we get
tan5x+tan3x = (sin5x•cos3x + cos5x•sin3x)/(cos3x•cos5x)
Doesn't the numerator on the right side of equation reminds something...
Well,
sin(a+b) = sina•cosb + sinb•cosa
Yep, put a=5 and b=3
So, we can rewrite the tangent s as,
tan5x+tan3x = sin(3x+5x)/(cos3x•cos5x)
Which is sin8x/(cos3x•cos5x)
Similarly with subtraction of 1) with 2) and using the equations
A/B-C/D = (AD-CB)/BD and
sin(a-b) = sina•cosb-sinb•cosa
2) tan3x = sin3x/cos3x
Add 1) and 2) and remember
A/B + C/D = (AD+BC)/(BD)
Using this relation we get
tan5x+tan3x = (sin5x•cos3x + cos5x•sin3x)/(cos3x•cos5x)
Doesn't the numerator on the right side of equation reminds something...
Well,
sin(a+b) = sina•cosb + sinb•cosa
Yep, put a=5 and b=3
So, we can rewrite the tangent s as,
tan5x+tan3x = sin(3x+5x)/(cos3x•cos5x)
Which is sin8x/(cos3x•cos5x)
Similarly with subtraction of 1) with 2) and using the equations
A/B-C/D = (AD-CB)/BD and
sin(a-b) = sina•cosb-sinb•cosa
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