tan b = 3 (Q. III); sin a = 2/3 (Q.I) cos^2 b = 1/(1 + tan^2 b) = 1/(1+ 9) = 1/10 -> cos b = - 0.32 sin^2 b = 1 - 1/10 = 9/10 cos ^a = 1 - sin^2 a = 1 - 4/9 = 5/9 ...

Feb 28, 2017 · Explanation: Use these trig identities: cos (a + b) = cos a.cos b - sin a.sin b cos (a - b) = cos a.cos b + sin a.sin b) In this case: # (cos a.cos b)^2 - (sin a.sin b)^2 = # #= (cos a.cos b - sin …

ptan (a)+qtan (b)= (p+q)tan { (a+b)/2} =>ptan (a)+qtan (b)= (p+q) (2sin ( (a+b)/2)cos ( (a-b)/2))/ (2cos ( (a+b)/2)cos ( (a-b)/2)) =>ptan (a)+qtan (b)= (p+q) (sin (a ...

#||A||=sqrt (4+49+1)" ;" ||A||=sqrt54# #"find :"||B||" magnitude of B"# #||B||=sqrt (B_x^2+B_y^2+B_z^2)# #||B||=sqrt (3^2+2^2+ (-8)^2)=sqrt (9+4+64)# #||B||=sqrt77# #cos alpha= (A*B)/ (||A||*||B||)# #cos …

Area = 935pi The chord and two radii, each drawn from the center to its respective end of the chord form an isosceles triangle. The angle, theta between the two radii is: theta = pi/6 - pi/8 theta = pi/24 let …

After applying the above formula, we take out multiples of 2pi from the trig arguments. Other identities used: sin (x+pi/2)=-cos (x) as well as cos 2 x = 2 cos ^2 x - 1.

A circle has a chord that goes from #pi/3 # to #pi/8 # radians on the circle. If the area of the circle is #25 pi #, what is the length of the chord? GeometryCirclesCircle Arcs and Sectors

to have: #int (1-cos (x/3))/sin (x/2)dx = 12 int sin^2t/ ( 3sint-4sin^3t) dt# simplifying: #int (1-cos (x/3))/sin (x/2)dx = 12 int sint/ ( 3-4sin^2t) dt# Now write the denominator as: #3-4sin^2t = 4 -4sin^2t -1 = 4 (1 …

See the proof below Rolle's Theorem states: If f is continuous on [a,b] and differentiable on (a,b), and if f (a)=f (b)=0, then there is some c in the interval (a,b ...

Using the fundamental relation sin^2+cos^2=1, we can cosines from sines: sin^2 (a)+cos^2 (a)=1 \implies cos (a) = sqrt (1-sin^2 (a)) (we're taking the positive root because we know that alla angles …