How do I solve this trigonometric equation?
2 sin^2a +sin a cos a(cot a - tan a)
- lenpol7Lv 71 month ago
CotA - Tan A = CosA / SinA - SinA/CosA
Cos^2A - Sin^2A /(CosASinA)
2Sin^2A + SinACosA( Cos^2A - Sin^2A) / (SinACosA)
Cancel down 'SinACosA
2Sin^2A + ( Cos^A - Sin^2A)
Use the Trig. Identity Cos^2A = 1 - Sin^2A
2Sin^2A + ( 1 - Sin^2A - Sin^2A)
2Sin^2A + 1 - 2Sin^2A = '1' NB ( 2Sin^2A add to 'zero'.
- PinkgreenLv 71 month ago
Mistake: what you posted was not an equation, after
making correction & re-post it.
- 1 month ago
i feel some thing is rong
- husoskiLv 72 months ago
Step 1 is to realize that it's not an equation. It's an expression. There's nothing to solve for. The value of a can be anythign that makes both (tan a) and (cot a) defined.
You can simplify, though. If I don't see an identity I can use right away, I find it helps to write everything in terms of sines and cosines:
2 sin² a + (sin a)(cos a)(cot a - tan a) = 2 sin² a + (sin a)(cos a)[(cos a)/(sin a) - (sin a)/(cos a)]
= 2 sin² a + (cos² a - sin² a)
That last step multiplied (sin a)(cos a) into each term in  brackets and canceled denominators. Finish with:
= sin² a + cos² a
That's only true when both cos a and sin a are nonzero, making both the original expression and the cancellation of (sin a)/(sin a) and (cos a)/(cos a) valid.
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- fcas80Lv 72 months ago
It's not an equation.
However, if you set it to zero, i think there are no solutions.
- nyphdinmdLv 72 months ago
work out the second term
sin(a)cos(a)(cot(a) - tan(a)) = sin(a)cos(a)( cos(a)/sin(a) - sin(a)/cos(a))
= cos^2(a) - sin^2(a)
2 sin^2a +sin a cos a(cot a - tan a) = 2*sin^2(a) + cos^2(a) - sin^2(a) = sin^2(a) +cos^2(a) = 1
- 2 months ago
I meant "calculate"
- rotchmLv 72 months ago
You can't solve it because there's nothing to solve. To "solve" an equation you must have an equals sign.
If you meant to find the roots then here just factor out sin(a).
What does this imply?
To your update. Still your question doesn't make sense. Calculate what? What you type is like 5 + x, then you say calculate 5+x... well, its 5+x.
Type your question exactly as is it is posed to you. Provide a pic if need be.