How to solve this?

Question

if sinA+cosA = p what's the value of tanA (interms of p).
Also, for what values of p the above equation becomes in-consistent?

Answer 1

sinA+cosA = p ................(1)
or, 1 + 2sinA.cosA = p^2 (squaring both the side)
or, 2sinA.cosA = p^2 - 1
Now, (sinA - cosA)^2 = 1 - 2sinA.cosA
= 1 - (p^2 - 1) = 2 - p^2
or, sinA - cosA = (+/-)sqrt(2 - p^2) .................(2)
Depending on the value of A,
where sinA > cosA, sinA - cosA = sqrt(2 - p^2) and
where sinA < cosA,
sinA - cosA = -sqrt(2 - p^2).
This is clear from sine and cosine graph.
So, from equation(1) and (2) and angle A
2sinA = p + sqrt(2 - p^2) and 2cosA = p - sqrt(2 - p^2)
OR, 2sinA = p - sqrt(2 - p^2) and 2cosA = p + sqrt(2 - p^2)
hence tanA = [p + sqrt(2 - p^2)] / [p - sqrt(2 - p^2)]
OR tanA = [p - sqrt(2 - p^2)] / [p + sqrt(2 - p^2)]

The equation is consistent for any value of A if p^2 <= 2
or, -sqrt(2) <= p <= sqrt(2)

Answer 2

A can only be -90 degrees to +90 degrees

since

TanA = SinA/CosA

This has infinite No of solutions.

For fun Take A = 45 degrees.

Then P = 2

And Tan A = 1

Answer 3

Sin A + cos A =p

multiply both sides by sqrt(1+1) = sqrt(2)

cos pi/4 Sin A + sin pi/4 cos A = p/sqrt(2)

sin(A+pi/4) = p/Sqrt(2)
the rhs should be between -1 ad 1

so p should be between -sqrt(2) and sqrt(2)
Now LHS =
cosA(1+tan A) = p
square boh sides
cos^2A(1+tan A)^2 = p
or(1+tanA)^2/(1+tan^A) =p
let tan A = x
(1+x)^2 = p(1+x^2)
or 1 + 2x + x^2 = p + p x^2
or x^(p-1)- 2x +(p-1) = 0

this is quadratic and can be solved for x which is Tan A

Answer 4

sinA+cosA = p
Squaring both side
sin^2 A + cos^2 A +2sin A cos A = p^2
1 +2sin A cos A = p^2
2sin A cos A = p^2 -1
sin 2A = p^2 -1
2 tan A/(1+tan^2 A)= p^2 -1
Reverse it
(1+tan^2 A)/2tan A = 1/(p^2 -1)
Applying component's and dividend
(1+tan^2 A+ 2tan A)/ (1+tan^2 A- 2tan A)= {1+(p^2 -1)}/{1-(p^2 -1)}
[(1+tan A)/ (1- tan A)]^2= {p^2 )}/{2-p^2}
Taking sq rt
[(1+tan A)/ (1- tan A)]= {p )}/sq rt {2-p^2}
Again apply components and dividend

[(1+tan A)+(1- tan A)]/ [(1+tan A)-(1- tan A)]= {p + sqr (2-p^2}/{p -sqrt {2-p^2}
[2]/ [2tan A)]= {p + sqr (2-p^2}/{p -sqrt {2-p^2}
1/ tan A= {p + sqr (2-p^2}/{p -sqrt {2-p^2}
Inverse it
tan A= {p - sqr (2-p^2}/{p +sqrt {2-p^2}

Answer 5

If p>sqrt2 (thank you Kaiser Wilhelm) then there is no value of A which would satisfy the equation because sin and cos are periodic functions.

By squaring both sides of the equation - you get sin2A = p-1. You would need to use identities to find tanA in terms of p but that problem is doable and looks like one poster already succeeded.

The previous poster's logic (who since removed her answer) is faulty because p cannot be 2 or -2 because there is no value of A where that would hold true. The previous poster could use a refresher course in trig. I need a refresher course as well and she can do my homework and give me an Easy A like she does for her students....

Answer 6

TanA = p/CosA - 1

the equation is inconsistent when p = SinA,
since TanA cannot equal TanA - 1.

By the way, who are you math freaks with all those equations? ; )

Answer 7

Orthodox, your logic is faulty is well. True f(a)=sin(a)+cos(a) doesn't acheive a value of 2 for any (real) a BUT one is not the upper bound either.

f(pi/4)=sqrt(2)=1.414...which is certainly bigger than one. You can certainliy check that by taking the derivative and setting it to zero which gives you sin(a)=cos(a). The minimum behaves the same way.

Answer 8

krs & amarsinghson gave you the correct answer for tan(A).

we can also express the answer as:

tan(A) = [ 1 - p√(2 - p²) ] ∕ ( p² - 1)

so the restrictions on p are:

|p| > 1 (from the triangle inequality)

and

2 - p² ≥ 0
-p² ≥ -2
p² ≤ 2
p ≤ ±√2

in total:

-√2 < p ≤ √2

my mistake i forgot the absolute value in the triangle ineq.

Answer 9

please read carefully as there are many brackets used
sina+cosa=p
divide and multiply by2^1/2
2^1/2(sina/2^1/2+cosa/2^1/2)=p
now 2^1/2=sin45=cos45
2^1/2(sinacos45+cosasin45)=p
2^1/2(sin(a+45)) =p
sin(a+45)=p/(2^1/2)
tan(a+45)=p/(2-(p^2))^1/2
expand tan(a+b)
tana+1/1-tana=p/(2-(p^2))^1/2
cross multiply
tana=(p-(2-(p^2))^1/2)/(p+(2-(p^2))^1/2)

Answer 10

I am learning tan, sin and cos in my maths class. I hate it... So tough.