Tricky Trig
sin θ + cos θ = 2 sin ( 9 0 ∘ − θ ) , cot θ = ?
Give your answer to the above problem to 3 decimal places.
The answer is 2.414.
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12 solutions
Thank you :)
How can you divide by a variable? what if cos x is 0?
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You can easily prove that cos x = 0 . Let's say if otherwise cos x = 0 , substitute it into the original equation you get sin x = 0 , however sin 2 x + cos 2 x = 0 = 1 , this is a contradiction. So cos x = 0 .
Whats wrong here ? Sin(x)cos(45) + cos(x)sin(45) =sin(90-x) Sin(45+x)=sin(90-x) Or 2x=45 X=45/2 Cot(45/2)= 1.7926
Done with same approach.. Best of luck :)
Looks easy but when you tries to solve, it becomes hard!!! but once again it becomes easy when you looks at the solution!!!
I got the right ans. i.e sqrt2 +1. But i approxed it to 2.41 just two decimal places and its says your ans. Is incorrect. Wth.
sin θ + cos θ = 2 sin ( 9 0 ° − θ )
sin θ + cos θ = 2 cos θ
sin θ = ( 2 − 1 ) cos θ
2 − 1 1 = sin θ cos θ
1 2 + 1 = cot θ
= > cot θ = 1 . 4 1 4 . . . + 1
cot θ ≈ 2 . 4 1 4
sin x + cos x=2^1/2 sin (90-x)
because of sin (90-x)= cos x,
sin x + cos x=2^1/2 cos x
dividing terms with sin x,
sin x/sin x + cos x/sin x=2^1/2 cos x/sin x
1+cot x=2^1/2 cot x
1= (2^1/2-1) cot x
cot x =2.41
Whats wrong here ? Sin(x)cos(45) + cos(x)sin(45) =sin(90-x) Sin(45+x)=sin(90-x) Or 2x=45 X=45/2 Cot(45/2)= 1.7926
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What's wrong is that you can't jump from sin ( 4 5 ∘ + x ) = sin ( 9 0 ∘ − x ) to 4 5 ∘ + x = 9 0 ∘ − x .
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so whats the right step after that ? please tell me ! thanks !
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@Purushottam Gupta – From there, you could take sin ( 9 0 ∘ − x ) to the RHS and simplify, but I don't see any efficient method from there.
Whats wrong here ? Sin(x)cos(45) + cos(x)sin(45) =sin(90-x) Sin(45+x)=sin(90-x) Or 2x=45 X=45/2 Cot(45/2)= 1.7926
Given ,
sin x + cos x = root(2) sin (90 - x) = root(2) cos x
sin x = root(2) cos x - cos x
Taking out cos x common from the RHS we get ,
sin x = cos x ( root(2) - 1)
sin x / cos x = root(2) - 1
tan x = 0.4142
cot x = 1 / 0.4142 = 2.414
Hence the value of cot x is 2.414 upto three decimal places.
Easy question if you have absolutely read trigonometry.
s i n θ + c o s θ = 2 c o s θ Dividing by s i n θ we get, 1 + c o t θ = 2 c o t θ Simplifying yields, c o t θ = 2 − 1 1
Does this question require a solution? It's too easy. No offence
sinx + cosx = sqrt(2)cosx
sinx={sqrt(2)-1}cosx
cosx/sinx=1/{sqrt(2)-1}
cotx=sqrt(2) + 1
So, cotx=2.414
Using r-formula, sinX + cosX = sqrt(2)sin(X + 45deg)
X + 45deg = 90deg - X
Thus one possible value of X is 67.5deg
Unfortunately, I got stuck here and had to use the calculator to find cot(67.5deg).
Esy.... sinx+cosx=2^1/2sin(90_x) =sinx+cosx=2^1/2cosx =sinx=(2^1/2-1)cosx =tanx=2^1/2-1 =cotx=2^1/2+1 =cotx=2.414 (approx)
sin ( 9 0 − x ) = cos ( x )
Using this identity
sin ( x ) + cos ( x ) = 2 cos ( x )
cos ( x ) sin ( x ) + cos ( x ) = 2
cos ( x ) sin ( x ) + cos ( x ) cos ( x ) = 2
Remember
cos ( x ) sin ( x ) = tan ( x )
∴ tan ( x ) + 1 = 2
tan ( x ) = 2 − 1
cot ( x ) = 2 − 1 1 ≈ 2 . 4 1 4