To calculate the equation COS tita = SIN 40, we need to find the value of tita that satisfies the equation.
Since COS tita = SIN 90 - tita, we can rewrite the equation as SIN 90 - tita = SIN 40.
Using the trigonometric identity SIN (90 - tita) = COS tita, we get COS tita = SIN 40.
Now we can rewrite the equation as COS tita = SIN (90 - 40).
Solving for tita:
COS tita = SIN 50
COS tita = SIN (90 - 40)
COS tita = SIN 50
tita = 90 - 50
tita = 40
Therefore, the value of tita that satisfies the equation COS tita = SIN 40 is 40 degrees.
Calculate equation
COS tita =SIN 40
9 answers
What is the symbol of Tita
The symbol for the variable "tita" is typically represented as θ (lowercase theta) in mathematics. It is commonly used to represent an angle in trigonometry and geometry.
Solve the equation
SIN Tita= COS 2 Tita
SIN Tita= COS 2 Tita
To solve the equation SIN tita = COS 2tita, we will use trigonometric identities to simplify and find the solutions.
Given: SIN tita = COS 2tita
Using the trigonometric identity COS 2tita = 1 - 2SIN^2 tita, we can rewrite the equation as:
SIN tita = 1 - 2SIN^2 tita
Rearranging terms, we get:
2SIN^2 tita + SIN tita - 1 = 0
Let y = SIN tita.
Therefore, the equation becomes:
2y^2 + y - 1 = 0
This is a quadratic equation in the form of ay^2 + by + c = 0.
Solving the quadratic equation using the quadratic formula y = (-b ± √(b^2 - 4ac)) / 2a, we find the values of y (SIN tita).
a = 2, b = 1, c = -1
y = (-1 ± √(1^2 - 4(2)(-1))) / 2(2)
y = (-1 ± √(1 + 8)) / 4
y = (-1 ± √9) / 4
y = (-1 ± 3) / 4
Therefore, the two possible solutions for SIN tita are:
1) y = ( -1 + 3) / 4 = 2 / 4 = 0.5
2) y = ( -1 - 3 ) / 4 = -4 / 4 = -1
However, the SIN function is restricted to the range [-1, 1], so the solution y = -1 is not valid.
Therefore, the solution for the equation SIN tita = COS 2tita is SIN tita = 0.5.
To find the value of tita:
Since SIN tita = 0.5, tita = 30 degrees or tita = π/6 radians.
Given: SIN tita = COS 2tita
Using the trigonometric identity COS 2tita = 1 - 2SIN^2 tita, we can rewrite the equation as:
SIN tita = 1 - 2SIN^2 tita
Rearranging terms, we get:
2SIN^2 tita + SIN tita - 1 = 0
Let y = SIN tita.
Therefore, the equation becomes:
2y^2 + y - 1 = 0
This is a quadratic equation in the form of ay^2 + by + c = 0.
Solving the quadratic equation using the quadratic formula y = (-b ± √(b^2 - 4ac)) / 2a, we find the values of y (SIN tita).
a = 2, b = 1, c = -1
y = (-1 ± √(1^2 - 4(2)(-1))) / 2(2)
y = (-1 ± √(1 + 8)) / 4
y = (-1 ± √9) / 4
y = (-1 ± 3) / 4
Therefore, the two possible solutions for SIN tita are:
1) y = ( -1 + 3) / 4 = 2 / 4 = 0.5
2) y = ( -1 - 3 ) / 4 = -4 / 4 = -1
However, the SIN function is restricted to the range [-1, 1], so the solution y = -1 is not valid.
Therefore, the solution for the equation SIN tita = COS 2tita is SIN tita = 0.5.
To find the value of tita:
Since SIN tita = 0.5, tita = 30 degrees or tita = π/6 radians.
Please use a simple method to solve the equation I don't understand the method you are using to solve use an easier method for me to understand
Absolutely! Let me provide a simpler method to solve the equation SIN tita = COS 2tita.
Given equation: SIN tita = COS 2tita
Using trigonometric identities, we know that COS 2tita = 1 - 2SIN^2 tita.
Substitute this into the equation:
SIN tita = 1 - 2SIN^2 tita
Rearrange the terms:
2SIN^2 tita + SIN tita - 1 = 0
This is a quadratic equation in SIN tita which can be rewritten as:
(2SIN tita - 1)(SIN tita + 1) = 0
From this, we get two solutions:
1) 2SIN tita - 1 = 0
2SIN tita = 1
SIN tita = 1/2
tita = 30 degrees or π/6 radians
2) SIN tita + 1 = 0
SIN tita = -1
However, the SIN function is restricted to the range [-1, 1], so the solution SIN tita = -1 is not valid.
Therefore, the solution to the equation SIN tita = COS 2tita is tita = 30 degrees or tita = π/6 radians.
Given equation: SIN tita = COS 2tita
Using trigonometric identities, we know that COS 2tita = 1 - 2SIN^2 tita.
Substitute this into the equation:
SIN tita = 1 - 2SIN^2 tita
Rearrange the terms:
2SIN^2 tita + SIN tita - 1 = 0
This is a quadratic equation in SIN tita which can be rewritten as:
(2SIN tita - 1)(SIN tita + 1) = 0
From this, we get two solutions:
1) 2SIN tita - 1 = 0
2SIN tita = 1
SIN tita = 1/2
tita = 30 degrees or π/6 radians
2) SIN tita + 1 = 0
SIN tita = -1
However, the SIN function is restricted to the range [-1, 1], so the solution SIN tita = -1 is not valid.
Therefore, the solution to the equation SIN tita = COS 2tita is tita = 30 degrees or tita = π/6 radians.
I don't understand
I'll simplify it further:
Given equation: SIN tita = COS 2tita
1. Using trigonometric identity: COS 2tita = 1 - 2SIN^2 tita
2. Substitute into the equation: SIN tita = 1 - 2SIN^2 tita
3. Rearrange terms to form a quadratic equation: 2SIN^2 tita + SIN tita - 1 = 0
4. Factor the quadratic equation: (2SIN tita - 1)(SIN tita + 1) = 0
5. Solve for SIN tita:
a) 2SIN tita - 1 = 0
2SIN tita = 1
SIN tita = 1/2
tita = 30 degrees or tita = π/6 radians
b) SIN tita + 1 = 0
SIN tita = -1 (not valid due to SIN function range)
Therefore, the solution to the equation SIN tita = COS 2tita is tita = 30 degrees or tita = π/6 radians.
Given equation: SIN tita = COS 2tita
1. Using trigonometric identity: COS 2tita = 1 - 2SIN^2 tita
2. Substitute into the equation: SIN tita = 1 - 2SIN^2 tita
3. Rearrange terms to form a quadratic equation: 2SIN^2 tita + SIN tita - 1 = 0
4. Factor the quadratic equation: (2SIN tita - 1)(SIN tita + 1) = 0
5. Solve for SIN tita:
a) 2SIN tita - 1 = 0
2SIN tita = 1
SIN tita = 1/2
tita = 30 degrees or tita = π/6 radians
b) SIN tita + 1 = 0
SIN tita = -1 (not valid due to SIN function range)
Therefore, the solution to the equation SIN tita = COS 2tita is tita = 30 degrees or tita = π/6 radians.
3 answers