Question from http://sgforums.com/forums/2297/topics/344664
It is given that x and y are acute angles such that sin(x-y) = 4/5 and sin x cos y = 9/10.
Show that
a)
(i) sin(x+y) = 1
(ii) tan x = 9 tan y
b) using the answers from part (a), find the value of tan x.
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Answer:
sin(x-y) = 4/5
sin x cos y - cos x sin y = 4/5
9/10 - cos x sin y = 4/5
cos x sin y = 1/10
a)
(i) sin (x + y)
= sin x cos y + cos x sin y
= 9/10 + 1/10
= 1 (shown)
(ii) (sin x cos y) / (cos x sin y) = (9/10) / (1/10)
(sin x / cos x) / (sin y / cos y) = 9
tan x / tan y = 9
tan x = 9 tan y (shown)
b) From sin (x+y) = 1, x + y = 90 degrees since x and y are acute.
Hence, tan (x+y) = infinity
tan (x+y) = (tan x + tan y) / (1 - tan x tan y)
===> tan x tan y = 1
===> tan y = 1 / tan x
Using a)(ii), tan x = 9 tan y
Hence,
tan x = 9 / tan x
tan² x = 9
tan x = sqrt (9)
tan x = 3 (since x is acute, tan is positive)
