Can someone please solve them,I just want to see examples,i will do the next 10 myself!!THX!
1)given that tan θ = 2/3 and that θ is acute,find the exact value of:
a)sin θ
b)cos θ
c)sin^2 θ
2)given that sin given that tan θ = (√2)/5 and that θ is acute,find the exact value of:
d)sin θ + cos θ
e)(cos θ - tan θ) / (1 - cos^2 θ)
1)given that tan θ = 2/3 and that θ is acute,find the exact value of:
a)sin θ
b)cos θ
c)sin^2 θ
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Draw that right triangle
opposite = 2
adjacent = 3
so hypotenuse = sqrt(4+9) = sqrt 13
sin = opposite / hyp = 2/sqrt 13
cos = adjacent / hyp = 3/sqrt 13
sin^2 = 4/13
2)given that [ sin given that <---- huh ? ] tan θ = (√2)/5 and that θ is acute,find the exact value of:
d)sin θ + cos θ
e)(cos θ - tan θ) / (1 - cos^2 θ)
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opposite = sqrt 2
adjacent = 5
so hypotenuse = sqrt (27)
sin = sqrt(2/27)
cos = sqrt (25/27)
etc remember 1 - cos^2 = sin^2
Given: X = 3, Y = 2.
a. r^2 = x^2 + y^2 = 3^2 + 2^2 = 13
r = sqrt(13).
sin A = y/r = 2/sqrt(13).
b. Cos A = x/r = 3/sqrt(13).
c.
2. Given: X = 5, Y = sqrt(2).
r^2 = x^2 + y^2 = 5^2 + (sqrt(2))^2 = 27
r = 3sqrt(3).
d. sin A + cos A = sqrt(2)/3sqrt(3) + 5/3sqrt(3) = (sqrt(2)+5)/3sqrt(3).
Sure! I can help you solve these examples and explain the steps to you.
1) Given that tan θ = 2/3 and θ is acute, let's find the exact values of:
a) sin θ:
To find sin θ, we can use the Pythagorean identity: sin^2 θ + cos^2 θ = 1.
Since tan θ = opposite/adjacent = 2/3, we can assign the opposite side as 2 and the adjacent side as 3.
Using the Pythagorean theorem, we can find the hypotenuse:
c^2 = a^2 + b^2
c^2 = 2^2 + 3^2
c^2 = 4 + 9
c^2 = 13
c = sqrt(13)
Now we can find sin θ:
sin θ = opposite/hypotenuse = 2/sqrt(13)
b) cos θ:
cos θ = adjacent/hypotenuse = 3/sqrt(13)
c) sin^2 θ:
sin^2 θ = (2/sqrt(13))^2 = 4/13
2) Given that tan θ = (√2)/5 and θ is acute, let's find the exact values of:
d) sin θ + cos θ:
To find sin θ and cos θ, we can use the Pythagorean identity sin^2 θ + cos^2 θ = 1.
Since tan θ = opposite/adjacent = (√2)/5, we can assign the opposite side as √2 and the adjacent side as 5.
Using the Pythagorean theorem, we can find the hypotenuse:
c^2 = a^2 + b^2
c^2 = (√2)^2 + 5^2
c^2 = 2 + 25
c^2 = 27
c = sqrt(27) [Note: We can simplify the answer as sqrt(27) = 3√3]
Now we can find sin θ and cos θ:
sin θ = opposite/hypotenuse = (√2)/sqrt(27) = (√2)/(3√3) = (√2)/(3√3) * (√3)/(√3) = (√6)/(3√9) = (√6)/3
cos θ = adjacent/hypotenuse = 5/sqrt(27) = 5/(3√3)
Now we can find sin θ + cos θ:
sin θ + cos θ = (√6)/3 + 5/(3√3)
To combine the terms, we multiply the numerator and denominator of the second term by √3 to eliminate the square root in the denominator:
sin θ + cos θ = (√6)/3 + 5/(3√3) * √3/√3 = (√6)/3 + (5√3)/(3√3) = (√6 + 5√3)/3
e) (cos θ - tan θ) / (1 - cos^2 θ):
Substituting the values we found earlier:
(cos θ - tan θ) / (1 - cos^2 θ) = (5/(3√3) - (√2)/5) / (1 - (5/(3√3))^2)
Simplify further as needed.
I hope this helps! Let me know if you have any more questions.