Trigonometric Identities - Miami
2 sin x y 2 The Law of Sines sinA a = sinB b = sinC c Suppose you are given two sides, a;band the angle Aopposite the side A The height of the triangle is h= bsinA Then
Formulas from Trigonometry
sin 2A+cos A= 1 sin(A B) = sinAcosB cosAsinB cos(A B) = cosAcosB tansinAsinB tan(A B) = A tanB 1 tanAtanB sin2A= 2sinAcosA cos2A= cos2 A sin2 A tan2A= 2tanA 1 2tan A sin A 2 = q 1 cosA 2 cos A 2 = q 1+cos A 2 tan 2 = sinA 1+cosA sin2 A= 1 2 21 2 cos2A cos A= 1 2 + 1 2 cos2A sinA+sinB= 2sin 1 2 (A+B)cos 1 2 (A 1B) sinA sinB= 2cos 1 2 (A+B)sin 2
TRIGONOMETRY LAWS AND IDENTITIES
TRIGONOMETRY LAWS AND IDENTITIES DEFINITIONS sin(x)= Opposite Hypotenuse cos(x)= Adjacent Hypotenuse tan(x)= Opposite Adjacent csc(x)= Hypotenuse Opposite sec(x)= Hypotenuse Adjacent
Trigonometric Identities
2 cos A B 2 (13) cosA cosB= 2sin A+ B 2 sin A B 2 (14) sinA+ sinB= 2sin A+ B 2 cos A B 2 (15) sinA sinB= 2cos A+ B 2 sin A B 2 (16) Note that (13) and (14) come from (4) and (5) (to get (13), use (4) to expand cosA= cos(A+ B 2 + 2) and (5) to expand cosB= cos(A+B 2 2), and add the results) Similarly (15) and (16) come from (6) and (7)
Domain, Range, and Period of the three main trigonometric
I sin(sin 1(x)) = xwhen 1 x 1 II sin 1(sin(x)) = xwhen ˇ 2 x ˇ 2 Good I sin(sin 1(1=2)) = 1=2, since 1 1=2 1 Bad I sin(sin 1( 1:8)) = unde ned, since 1:8 < 1 Good II: is in the right quadrant, and written correctly sin 1(sin ˇ 5 ) = ˇ 5, since ˇ 2 ˇ 5 ˇ 2 Bad II: is in the right quadrant, but written incorrectly sin 1(sin 9ˇ 5
AP CALCULUS BC 2011 SCORING GUIDELINES
Let f ()xx x=+sin cos ()2 The graph of yf x= ()5 is shown above (a) Write the first four nonzero terms of the Taylor series for sin x about x =0, and write the first four nonzero terms of the Taylor series for sin()x2 about x =0 (b) Write the first four nonzero terms of the Taylor series for cos x about x =0 Use this series and the series for
Table of Integrals
sin[2(a+ b)x] 16(a+ b) (76) Z sin2 axcos2 axdx= x 8 sin4ax 32a (77) Z tanaxdx= 1 a lncosax (78) Z tan2 axdx= x+ 1 a tanax (79) Z tann axdx= tann+1 ax a(1 + n) 2F 1 n+
The Squeeze Theorem: Statement and Example
x→0 sin(x) x =1: To do this, we’ll use the Squeeze theorem by establishing upper and lower bounds on sin(x)~x in an interval around 0 Speci cally, we’ll show that cos(x) ≤ sin(x) x ≤1 in an interval around 0 We can already see why this should be the case by the following graph y=cos(x) y= sin(x) x y=1 2
Math 2260 Exam Practice Problem Solutions
Answer: First, recall that the Taylor series centered at x= 0 for sin(x) is sin(x) = x x3 3 + x5 5 x7 7 + :::: Therefore, the Taylor series centered at x= 0 for sin(x2) is sin(x2) = (x2) (x2)3 3 + (x2)5 5 (x2)7 7 + :::= x2 x6 3 + x10 5 x14 7 + :::: Hence, Z 1 0 sin(x2)dx= Z 1 0 x2 x6 3 + x10 5::: dx= x3 3 x7 7 3 + x11 11 5::: 1 = 1
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