Trigonometric Ratios and Identities

Given that sinhx=‌

12

5

, we denote cosh‌x=t. We utilize the identity for hyperbolic functions: cosh2x− sinh2x=1
Substituting the known value, we have:
‌t2−(‌

12

5

)2=1
‌t2−‌

144

25

=1
Solving for t2, we get:
t2=‌

169

25

Thus, t=‌

13

5

.
Now, we need to find sinh3x+cosh‌3‌x. We use the identities for triple angles:
‌ sinh3x=3 sinhx+4 sinh3x
‌cosh‌3‌x=4cosh3x−3‌cosh‌x
So,
sinh3x+cosh‌3‌x=3 sinhx+4 sinh3x+4cosh3x−3‌cosh‌x
We substitute the values:
‌=3(‌

12

5

−‌

13

5

)+4(‌

12

5

+‌

13

5

)(‌

144

25

+‌

169

25

−‌

156

25

)
‌=3(‌12∕5−‌13∕5)+4(‌25∕5)(‌157∕25)
‌=‌

−3

5

+4⋅‌

157

25

‌=‌

−3

5

+‌

628

25

‌=‌

−3

5

+‌

628

25

=‌

−3

5

+‌

2512

125

‌=‌

−75+2512

125

=‌

2437

125

Thus, the calculated value is:
=125