Can You Solve 0.221=sinθcosθ for θ Without a Calculator?

  • Thread starter Thread starter hahaha158
  • Start date Start date
AI Thread Summary
To solve the equation 0.221 = sinθcosθ, one can utilize the identity sin(2θ) = 2sin(θ)cos(θ), which simplifies the problem. By rewriting the equation as 0.1105 = sin(2θ), it becomes easier to solve for θ. The next step involves finding the inverse sine of 0.1105, which can be done using a calculator or trigonometric tables. If a calculator is not available, one can estimate the angle using known values of sine. This approach provides a clear path to determine the value of θ without relying solely on computational tools.
hahaha158
Messages
79
Reaction score
0

Homework Statement



i have the equation 0.221=sinθcosθ. I want to find the vaue of θ. Is there anyway i can solve this with or without a calculator? (solver function on both my calculator and wolfram alpha do not work)

thanks
 
Physics news on Phys.org
hahaha158 said:

Homework Statement



i have the equation 0.221=sinθcosθ. I want to find the vaue of θ. Is there anyway i can solve this with or without a calculator? (solver function on both my calculator and wolfram alpha do not work)

thanks

sin(2θ)=2*sin(θ)*cos(θ). Try using that.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Mastering Double Angle Identities: Solving -2sin3θ+1=0
Replies
10
Views
2K
How do you solve cos(α+θ)=0 for different values of α?
Replies
21
Views
2K
Computing arctan (inverse tan) function
Replies
10
Views
3K
I Why does this concavity function not work for this polar fun
Replies
4
Views
2K
Solving a system of two simultaneous trigonometric equations
Replies
14
Views
3K
Solving trigonometric equations as fractions of π
Replies
5
Views
2K
Efficient Logarithmic Calculations for 0.3048 without a Calculator
Replies
8
Views
2K
Solve acos²θ+bsinθ+c=0 for all values 0≤θ≤360°
Replies
11
Views
2K
Why does [sin(θ/2)-cos(θ/2)]^2 not always equal 1?
Replies
8
Views
3K
Trigonometric Equations Problems - Rather Confused
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top