What is the exact value of sin when cos x =1/√10?

AI Thread Summary
To find the exact value of sin when cos x = 1/√10 for a first quadrant angle, one can use the Pythagorean identity sin²x + cos²x = 1. Substituting cos x into the equation gives sin²x + (1/√10)² = 1, leading to sin²x = 1 - 1/10, which simplifies to sin²x = 9/10. Taking the square root results in sin x = 3/√10, as sin must be positive in the first quadrant. Thus, the exact value of sin when cos x = 1/√10 is 3/√10.
ohhnana
Messages
25
Reaction score
0

Homework Statement



cos x is a first quadrant angle in standard position and cos x =1/√10 . Find the exact value of sin .

Homework Equations



?

The Attempt at a Solution


?
 
Physics news on Phys.org
Hint: Remember the definition of sine and cosine.
 
I would suggest drawing a triangle with the fraction result of cos x distributed on the appropriate edges. You should be able to 'fill in the blanks'
 
ohhnana said:

Homework Statement



cos x is a first quadrant angle in standard position and cos x =1/√10 . Find the exact value of sin .

Homework Equations



?

The Attempt at a Solution


?
Excerpt from the forum rules (https://www.physicsforums.com/showthread.php?t=414380)
Homework Help:
You MUST show that you have attempted to answer your question in order to receive help. You MUST make use of the homework template, which automatically appears when a new topic is created in the homework help forums.
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
View full post »

Similar threads

How to find out the sin value from cos
Replies
9
Views
2K
Solve the problem that involves ##\cos^{-1} x + \cos^{-1}y##
Replies
3
Views
1K
To find the range of a given ##\sin## function
Replies
15
Views
2K
Values of sin and cos (rad and deg)
Replies
4
Views
2K
Solving equations of the form ##a\sin\theta+b\cos\theta = c##
Replies
2
Views
2K
I know ##tan 2\theta## but what is ##sin \theta##
Replies
8
Views
2K
State the exact values of all the trigonometric rations for theta.
Replies
12
Views
2K
Finding the exact value of cos(arcsin())
Replies
33
Views
5K
Solve Trigonometry Problem: Find the Value of a Series of Cosine Squared Terms"
Replies
22
Views
4K
Proving Trigonometric Identities: (sin φ+1-cos φ)/(sin φ+cos φ-1)
Replies
12
Views
2K

Hot Threads

Recent Insights

Back
Top