Question

If (sin A + cos A)=7/5.
Find:
i)sin 2a
ii)cos 2a
iii)sec A + Cosec A

Answer 1

Is A = a?

Answer 2

\[\sin A + \cos A = \frac{7}{5}\]\[\left( \sin A + \cos A \right)^{2} = \frac{49}{25}\]\[\sin^{2} A + \cos^{2} A + 2 \sin A \cos A = \frac{49}{25}\]\[1 + \sin 2A = \frac{49}{25}\]\[\sin 2A = \frac{24}{25}\]draw a triangle to find cos 2A

Answer 3

Looks like a 3,4,5 triangle maybe....

Answer 4

i) sin2A=2sinAcosA=(sinA+cosA)^2 - (sin^2 A + cos^2 A)=(7/5)^2 -1=24/25
ii)
cos2A=7/25
iii)sec A + Cosec A =1/cosA + 1/sinA = (sinA+cosA)/sinAcosA
=2(sinA+cosA)/2sinAcosA = 2(7/5)/(24/25)=35/12

Answer 5

for iii\[\sec A + \csc A = \frac{1}{\cos A} + \frac{1}{\sin A}\]\[= \frac{\sin A + \cos A}{\sin A \cos A}\]\[= \frac{2(\sin A + \cos A)}{\sin 2A}\]

Answer 6

eg 3,4,5 for sin a + cos a is 3/5 + 4/5 and

sin 2a = 2 sin a cos a = 2*3/5*4/5 = 24/25

Answer 7

Thats awesome tysm @exraven ,and ty everyone else :)