tan(arctan(45))tan(arctan(45))

u = sin(v) cos(v) = sqrt(1-(sin(v))^2) = sqrt(1-u^2) But v = arcsin(u), therefore cos(arcsin(u)) = sqrt(1-u^2). Next replace u with (x-h)/r to get: cos(arcsin((x-h)/r)) = sqrt(1-(x-h)^2/r^2).

MATH

csc(arctan(x2–√))csc(arctan(x2)) Write an algebraic expression that is equivalent to the given…

Taking x as oppsite side and 2–√2 as the adjacent side,the hypotenuse will be equal to x2+2−−−−−√.x2+2. tan−1(x2–√)=csc−1(x2+2−−−−−√x)tan-1(x2)=csc-1(x2+2x)…

MATH

cot(arctan(1x))cot(arctan(1x)) Write an algebraic expression that is equivalent to the given expression….

Taking 1 as oppsite side and x as the adjacent side,the hypotenuse will be equal to 1+x2−−−−−√.1+x2. tan−1(1x)=cot−1(x1)=cot−1xtan-1(1x)=cot-1(x1)=cot-1x cot(tan−1(1x))=cot(cot−1x)=xcot(tan-1(1x))=cot(cot-1x)=x

MATH

tan(arccos(x3))tan(arccos(x3)) Write an algebraic expression that is equivalent to the given expression….

Taking x as adjacent side and 3 as the hypotenuse,adjacent side will be equal to 32−x2−−−−−−√=9−x2−−−−−√32-x2=9-x2 cos−1(x3)=tan−1(9−x2−−−−−√x)cos-1(x3)=tan-1(9-x2x) tan(cos−1(x3))=tan(tan−1(9−x2−−−−−√x))tan(cos-1(x3))=tan(tan-1(9-x2x))…

MATH

sec(arcsin(x−1))sec(arcsin(x-1)) Write an algebraic expression that is equivalent to the given…

Taking (x-1) as oppsite side and 1 as the hypotenuse,the adjacent side will be equal to 1−(x−1)2−−−−−−−−−−√=2x−x2−−−−−−√1-(x-1)2=2x-x2 sin−1(x−1)=sec−1(12x−x2−−−−−−√)sin-1(x-1)=sec-1(12x-x2)…

MATH

sin(arccos(x))sin(arccos(x)) Write an algebraic expression that is equivalent to the given expression….

sin(cos−1x)=sin(sin−1(1−x2−−−−−√))sin(cos-1x)=sin(sin-1(1-x2)) =1−x2−−−−−√=1-x2

MATH

sec(arctan(3x))sec(arctan(3x)) Write an algebraic expression that is equivalent to the given expression….

Taking 3x as oppsite side and 1 as the adjacent side,the hypotenuse will be equal to 9×2+1−−−−−−√.9×2+1. So tan−1(3x)=sec−1(9×2+1−−−−−−√1)tan-1(3x)=sec-1(9×2+11) sec(tan−1(3x))=sec(sec−1(9×2+1−−−−−−√))sec(tan-1(3x))=sec(sec-1(9×2+1))…

MATH

cos(arcsin(2x))cos(arcsin(2x)) Write an algebraic expression that is equivalent to the given expression….

cos(sin−1(2x))=cos(cos−1(1−(2x)2−−−−−−−−√))cos(sin-1(2x))=cos(cos-1(1-(2x)2)) =cos(cos−1(1−4×2−−−−−−√))=cos(cos-1(1-4×2)) =1−4×2−−−−−−√=1-4×2

MATH

sin(arctan(x))sin(arctan(x)) Write an algebraic expression that is equivalent to the given expression….

sin(arctan(x))sin(arctan(x)) let θ=arctan(x)θ=arctan(x) => tan(θ)=x=x1tan(θ)=x=x1 by using the right triangle we get the hypotenuse is =1+x2−−−−−√=1+x2 so sin(θ)=x1+x2−−−−−√sin(θ)=x1+x2

MATH

cot(arctan(x))cot(arctan(x)) Write an algebraic expression that is equivalent to the given expression….

tan−1(x)=cot−1(1x)tan-1(x)=cot-1(1x) cot(tan−1(x))=cot(cot−1(1x))=1xcot(tan-1(x))=cot(cot-1(1x))=1x

MATH

sec(sin−1(−2–√2))sec(sin-1(-22)) Find the exact value of the expression. (Hint: Sketch a right…

sec(sin−1(−2–√2))sec(sin-1(-22)) Letθ=sin−1(−2–√2)Letθ=sin-1(-22) so we need to find sec(θ)sec(θ) => sin(θ)=(−2–√2)sin(θ)=(-22) =>sin(θ)=−12–√sin(θ)=-12 we know

MATH

csc(cos−1(3–√2))csc(cos-1(32)) Find the exact value of the expression. (Hint: Sketch a right…

First, find the inner expression. Set cosx=−3–√2cosx=-32 because we need to find what value of x yields −3–√2-32 Use a unit circle to find this value => x=5π6x=5π6 Draw a triangle, Given the…

MATH

cot(arctan(58))cot(arctan(58)) Find the exact value of the expression. (Hint: Sketch a right triangle.)

cotx=1tanxcotx=1tanx Given, => cot(tan−1(58))cot(tan-1(58)) We know that cotx=1tanxcotx=1tanx Therefore, => 1tan(arctan(58))1tan(arctan(58)) tan and arctan cancel => 158158 Simplify => 8585

MATH

sin(arccos(−23))sin(arccos(-23)) Find the exact value of the expression. (Hint: Sketch a right triangle.)

We know that cos(arccos(−23))=−23cos(arccos(-23))=-23 Therefore, using trig identities, => sin=1−cos2−−−−−−−√=1−49−−−−−√sin=1-cos2=1-49 Thus, => sin=59−−√=5–√3sin=59=53

MATH

tan(arcsin(−34))tan(arcsin(-34)) Find the exact value of the expression. (Hint: Sketch a right triangle.)

tan(sin−1(−34))=−tan(sin−1(34))tan(sin-1(-34))=-tan(sin-1(34)) sin−1(34)=tan−1(37–√)sin-1(34)=tan-1(37) tan(sin−1(−34))=−tan(tan−1(37–√))=−37–√=−1.134tan(sin-1(-34))=-tan(tan-1(37))=-37=-1.134

MATH

sec(arctan(−35))sec(arctan(-35)) Find the exact value of the expression. (Hint: Sketch a right triangle.)

tan−1(−35)=−tan−1(35)tan-1(-35)=-tan-1(35) Taking 3 as opposite side and 5 as adjacent side , then the hypotenuse will be 52+32−−−−−−√=34−−√.52+32=34. tan−1(35)=sec−1(34−−√5)tan-1(35)=sec-1(345)

MATH

csc(arctan(−512))csc(arctan(-512)) Find the exact value of the expression. (Hint: Sketch a right triangle.)

csc(tan−1(−512))=−csc(tan−1(512))csc(tan-1(-512))=-csc(tan-1(512)) Taking 5 as opposite side and 12 as adjacent side ,then the hypotenuse will be 122+52−−−−−−−√=13.122+52=13. tan−1(512)=csc−1(135)tan-1(512)=csc-1(135)…

MATH

cos(arcsin(513))cos(arcsin(513)) Find the exact value of the expression. (Hint: Sketch a right triangle.)

Taking opposite side as 5 and hypotenuse as 13 ,then the adjacent side will be sqrt(13^2-5^2)=12. sin^-1(5/13)=cos^-1(12/13). cos(sin^-1(5/13))=cos(cos^-1(12/13))=12/13=0.923

MATH

sin(cos−1(5–√5))sin(cos-1(55)) Find the exact value of the expression. (Hint: Sketch a right…

5–√5=15–√55=15 Taking adjacent as 1 and hypotenuse as 5–√5 ,then the opposite will be (5–√)2−1−−−−−−−−√=2.(5)2-1=2. cos−1(15–√)=sin−1(25–√).cos-1(15)=sin-1(25)….

MATH

cos(tan−1(2))cos(tan-1(2)) Find the exact value of the expression. (Hint: Sketch a right triangle.)

Taking 2 as oppsite side and 1 as the adjacent side,the hypotenuse will be equal to 22+12−−−−−−√=5–√.22+12=5. tan−1(2)=cos−1(15–√)tan-1(2)=cos-1(15) cos(tan−1(2))=cos(cos−1(15–√))=15–√.cos(tan-1(2))=cos(cos-1(15))=15.

MATH

sec(arcsin(45))sec(arcsin(45)) Find the exact value of the expression. (Hint: Sketch a right triangle.)

Taking 4 as oppsite side and 5 as the hypotenuse ,adjacent side will be equal to 52−42−−−−−−√=3.52-42=3. sin−1(45)=sec−1(53)sin-1(45)=sec-1(53) sec(sin−1(45))=sec(sec−1(53))=53.sec(sin-1(45))=sec(sec-1(53))=53.

MATH

sin(arctan(34))sin(arctan(34)) Find the exact value of the expression. (Hint: Sketch a right triangle.)

Taking 3 as oppsite side and 4 as the adjacent side,the hypotenuse will be equal to 32+42−−−−−−√=5.32+42=5. tan−1(34)=sin−1(35)tan-1(34)=sin-1(35) sin(tan−1(34))=sin(sin−1(35))=35.sin(tan-1(34))=sin(sin-1(35))=35.

MATH

arccos(cos(7π2))arccos(cos(7π2)) Use the properties of inverse trigonometric functions to evaluate the…

arccos(cos(7π2))arccos(cos(7π2)) this could be thought of as 3 whole rounds (6π2)(6π2)plusπ2π2 as π2π2 is between [-1,1] we don’t need to do anything. the arccos is the inverse of cos, so…

MATH

arcsin(sin(3π))arcsin(sin(3π)) Use the properties of inverse trigonometric functions to evaluate the…

By the properties of inverse functions, a function and its inverse function will ‘cancel’ each other out and result in simply the input of the function. In this problem, we know that arcsinarcsin and…

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